Elementary Functions Part 1, Functions Lecture 1.1b, Functions - - PowerPoint PPT Presentation

Elementary Functions

Part 1, Functions Lecture 1.1b, Functions defined by equations

• Dr. Ken W. Smith

Sam Houston State University

2013

Smith (SHSU) Elementary Functions 2013 13 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Functions defined by equations

Many functions we explore in mathematics and science are defined by an equation. We can define a function implicitly in an equation involving two variables. For example, does the equation 2x + 3y − 4 = 0 define a function with inputs x and outputs y? Isolate y to get 3y = 4 − 2x and so y = 1 3(4 − 2x). We may now explicitly define the function f(x) = 1 3(4 − 2x) . So YES, the equation 2x + 3y − 4 = 0 does define a function.

Smith (SHSU) Elementary Functions 2013 14 / 27

Independent and dependent variables

A digression. When we considered the equation 2x + 3y − 4 = 0

• ur choice of x as input and y as output is arbitrary. We could decide (contrary to

custom!) that y is the input and x is the output. Then, solving for x, we have 2x = 4 − 3y and so x = 1 2(4 − 3y) and so we create the function g(y) = 1 2(4 − 3y). But most of the time we will stick to convention and, unless stated otherwise, assume x is the input variable and y is the output variable. The input variable x is often called the independent variable and the output variable is the dependent variable since its value depends on the input.

Smith (SHSU) Elementary Functions 2013 15 / 27

Independent and dependent variables

A digression. When we considered the equation 2x + 3y − 4 = 0

• ur choice of x as input and y as output is arbitrary. We could decide (contrary to

custom!) that y is the input and x is the output. Then, solving for x, we have 2x = 4 − 3y and so x = 1 2(4 − 3y) and so we create the function g(y) = 1 2(4 − 3y). But most of the time we will stick to convention and, unless stated otherwise, assume x is the input variable and y is the output variable. The input variable x is often called the independent variable and the output variable is the dependent variable since its value depends on the input.

Smith (SHSU) Elementary Functions 2013 15 / 27

Independent and dependent variables

A digression. When we considered the equation 2x + 3y − 4 = 0

• ur choice of x as input and y as output is arbitrary. We could decide (contrary to

custom!) that y is the input and x is the output. Then, solving for x, we have 2x = 4 − 3y and so x = 1 2(4 − 3y) and so we create the function g(y) = 1 2(4 − 3y). But most of the time we will stick to convention and, unless stated otherwise, assume x is the input variable and y is the output variable. The input variable x is often called the independent variable and the output variable is the dependent variable since its value depends on the input.

Smith (SHSU) Elementary Functions 2013 15 / 27

Independent and dependent variables

A digression. When we considered the equation 2x + 3y − 4 = 0

• ur choice of x as input and y as output is arbitrary. We could decide (contrary to

custom!) that y is the input and x is the output. Then, solving for x, we have 2x = 4 − 3y and so x = 1 2(4 − 3y) and so we create the function g(y) = 1 2(4 − 3y). But most of the time we will stick to convention and, unless stated otherwise, assume x is the input variable and y is the output variable. The input variable x is often called the independent variable and the output variable is the dependent variable since its value depends on the input.

Smith (SHSU) Elementary Functions 2013 15 / 27

Independent and dependent variables

A digression. When we considered the equation 2x + 3y − 4 = 0

• ur choice of x as input and y as output is arbitrary. We could decide (contrary to

custom!) that y is the input and x is the output. Then, solving for x, we have 2x = 4 − 3y and so x = 1 2(4 − 3y) and so we create the function g(y) = 1 2(4 − 3y). But most of the time we will stick to convention and, unless stated otherwise, assume x is the input variable and y is the output variable. The input variable x is often called the independent variable and the output variable is the dependent variable since its value depends on the input.

Smith (SHSU) Elementary Functions 2013 15 / 27

Independent and dependent variables

A digression. When we considered the equation 2x + 3y − 4 = 0

• ur choice of x as input and y as output is arbitrary. We could decide (contrary to

custom!) that y is the input and x is the output. Then, solving for x, we have 2x = 4 − 3y and so x = 1 2(4 − 3y) and so we create the function g(y) = 1 2(4 − 3y). But most of the time we will stick to convention and, unless stated otherwise, assume x is the input variable and y is the output variable. The input variable x is often called the independent variable and the output variable is the dependent variable since its value depends on the input.

Smith (SHSU) Elementary Functions 2013 15 / 27

Independent and dependent variables

A digression. When we considered the equation 2x + 3y − 4 = 0

• ur choice of x as input and y as output is arbitrary. We could decide (contrary to

custom!) that y is the input and x is the output. Then, solving for x, we have 2x = 4 − 3y and so x = 1 2(4 − 3y) and so we create the function g(y) = 1 2(4 − 3y). But most of the time we will stick to convention and, unless stated otherwise, assume x is the input variable and y is the output variable. The input variable x is often called the independent variable and the output variable is the dependent variable since its value depends on the input.

Smith (SHSU) Elementary Functions 2013 15 / 27

Independent and dependent variables

A digression. When we considered the equation 2x + 3y − 4 = 0

• ur choice of x as input and y as output is arbitrary. We could decide (contrary to

custom!) that y is the input and x is the output. Then, solving for x, we have 2x = 4 − 3y and so x = 1 2(4 − 3y) and so we create the function g(y) = 1 2(4 − 3y). But most of the time we will stick to convention and, unless stated otherwise, assume x is the input variable and y is the output variable. The input variable x is often called the independent variable and the output variable is the dependent variable since its value depends on the input.

Smith (SHSU) Elementary Functions 2013 15 / 27

Exercises on implicit functions

Some worked exercises.

1 Does the equation x2y = 4 define y as a function of x? (If it does, give the

domain of the implied function.)

• Solution. We attempt to solve for y. We may multiply both sides of the

equation by 1 x2 as long as x is not zero. This gives us y = 4 x2 . Is there a problem with x = 0? No, x = 0 does not allow x2y = 4, so x will never be zero in this equation. Answer: YES, this is a function; y = 4 x2 . The domain of this function is all real numbers except zero. In interval notation the domain is (−∞, 0) ∪ (0, ∞).

Smith (SHSU) Elementary Functions 2013 16 / 27

Exercises on implicit functions

Some worked exercises.

1 Does the equation x2y = 4 define y as a function of x? (If it does, give the

domain of the implied function.)

• Solution. We attempt to solve for y. We may multiply both sides of the

equation by 1 x2 as long as x is not zero. This gives us y = 4 x2 . Is there a problem with x = 0? No, x = 0 does not allow x2y = 4, so x will never be zero in this equation. Answer: YES, this is a function; y = 4 x2 . The domain of this function is all real numbers except zero. In interval notation the domain is (−∞, 0) ∪ (0, ∞).

Smith (SHSU) Elementary Functions 2013 16 / 27

Exercises on implicit functions

Some worked exercises.

1 Does the equation x2y = 4 define y as a function of x? (If it does, give the

domain of the implied function.)

• Solution. We attempt to solve for y. We may multiply both sides of the

equation by 1 x2 as long as x is not zero. This gives us y = 4 x2 . Is there a problem with x = 0? No, x = 0 does not allow x2y = 4, so x will never be zero in this equation. Answer: YES, this is a function; y = 4 x2 . The domain of this function is all real numbers except zero. In interval notation the domain is (−∞, 0) ∪ (0, ∞).

Smith (SHSU) Elementary Functions 2013 16 / 27

Exercises on implicit functions

Some worked exercises.

1 Does the equation x2y = 4 define y as a function of x? (If it does, give the

domain of the implied function.)

• Solution. We attempt to solve for y. We may multiply both sides of the

equation by 1 x2 as long as x is not zero. This gives us y = 4 x2 . Is there a problem with x = 0? No, x = 0 does not allow x2y = 4, so x will never be zero in this equation. Answer: YES, this is a function; y = 4 x2 . The domain of this function is all real numbers except zero. In interval notation the domain is (−∞, 0) ∪ (0, ∞).

Smith (SHSU) Elementary Functions 2013 16 / 27

Exercises on implicit functions

Some worked exercises.

1 Does the equation x2y = 4 define y as a function of x? (If it does, give the

domain of the implied function.)

• Solution. We attempt to solve for y. We may multiply both sides of the

equation by 1 x2 as long as x is not zero. This gives us y = 4 x2 . Is there a problem with x = 0? No, x = 0 does not allow x2y = 4, so x will never be zero in this equation. Answer: YES, this is a function; y = 4 x2 . The domain of this function is all real numbers except zero. In interval notation the domain is (−∞, 0) ∪ (0, ∞).

Smith (SHSU) Elementary Functions 2013 16 / 27

Exercises on implicit functions

Some worked exercises.

1 Does the equation x2y = 4 define y as a function of x? (If it does, give the

domain of the implied function.)

• Solution. We attempt to solve for y. We may multiply both sides of the

equation by 1 x2 as long as x is not zero. This gives us y = 4 x2 . Is there a problem with x = 0? No, x = 0 does not allow x2y = 4, so x will never be zero in this equation. Answer: YES, this is a function; y = 4 x2 . The domain of this function is all real numbers except zero. In interval notation the domain is (−∞, 0) ∪ (0, ∞).

Smith (SHSU) Elementary Functions 2013 16 / 27

Exercises on implicit functions

Some worked exercises.

1 Does the equation x2y = 4 define y as a function of x? (If it does, give the

domain of the implied function.)

• Solution. We attempt to solve for y. We may multiply both sides of the

equation by 1 x2 as long as x is not zero. This gives us y = 4 x2 . Is there a problem with x = 0? No, x = 0 does not allow x2y = 4, so x will never be zero in this equation. Answer: YES, this is a function; y = 4 x2 . The domain of this function is all real numbers except zero. In interval notation the domain is (−∞, 0) ∪ (0, ∞).

Smith (SHSU) Elementary Functions 2013 16 / 27

Exercises on implicit functions

Some worked exercises.

1 Does the equation x2y = 4 define y as a function of x? (If it does, give the

domain of the implied function.)

• Solution. We attempt to solve for y. We may multiply both sides of the

equation by 1 x2 as long as x is not zero. This gives us y = 4 x2 . Is there a problem with x = 0? No, x = 0 does not allow x2y = 4, so x will never be zero in this equation. Answer: YES, this is a function; y = 4 x2 . The domain of this function is all real numbers except zero. In interval notation the domain is (−∞, 0) ∪ (0, ∞).

Smith (SHSU) Elementary Functions 2013 16 / 27

Exercises on implicit functions

Some worked exercises.

1 Does the equation x2y = 4 define y as a function of x? (If it does, give the

domain of the implied function.)

• Solution. We attempt to solve for y. We may multiply both sides of the

equation by 1 x2 as long as x is not zero. This gives us y = 4 x2 . Is there a problem with x = 0? No, x = 0 does not allow x2y = 4, so x will never be zero in this equation. Answer: YES, this is a function; y = 4 x2 . The domain of this function is all real numbers except zero. In interval notation the domain is (−∞, 0) ∪ (0, ∞).

Smith (SHSU) Elementary Functions 2013 16 / 27

Not a function

2 Does the equation xy2 = 4 define y as a function of x?

• Solution. If we attempt to solve for y, we multiply both sides of the

equation by 1 x (as long as x = 0) and so we have y2 = 4 x. But now, what is y? y could be positive or negative – there will generally be two choices here, one positive and one negative. The appearance of two answers violates the uniqueness requirement in our

• utputs for a function.

Answer: NO, this is not a function. If x = 1 then we don’t know if y = 2 or y = −2.

Smith (SHSU) Elementary Functions 2013 17 / 27

Not a function

2 Does the equation xy2 = 4 define y as a function of x?

• Solution. If we attempt to solve for y, we multiply both sides of the

equation by 1 x (as long as x = 0) and so we have y2 = 4 x. But now, what is y? y could be positive or negative – there will generally be two choices here, one positive and one negative. The appearance of two answers violates the uniqueness requirement in our

• utputs for a function.

Answer: NO, this is not a function. If x = 1 then we don’t know if y = 2 or y = −2.

Smith (SHSU) Elementary Functions 2013 17 / 27

Not a function

2 Does the equation xy2 = 4 define y as a function of x?

• Solution. If we attempt to solve for y, we multiply both sides of the

equation by 1 x (as long as x = 0) and so we have y2 = 4 x. But now, what is y? y could be positive or negative – there will generally be two choices here, one positive and one negative. The appearance of two answers violates the uniqueness requirement in our

• utputs for a function.

Answer: NO, this is not a function. If x = 1 then we don’t know if y = 2 or y = −2.

Smith (SHSU) Elementary Functions 2013 17 / 27

Not a function

2 Does the equation xy2 = 4 define y as a function of x?

• Solution. If we attempt to solve for y, we multiply both sides of the

equation by 1 x (as long as x = 0) and so we have y2 = 4 x. But now, what is y? y could be positive or negative – there will generally be two choices here, one positive and one negative. The appearance of two answers violates the uniqueness requirement in our

• utputs for a function.

Answer: NO, this is not a function. If x = 1 then we don’t know if y = 2 or y = −2.

Smith (SHSU) Elementary Functions 2013 17 / 27

Not a function

2 Does the equation xy2 = 4 define y as a function of x?

• Solution. If we attempt to solve for y, we multiply both sides of the

equation by 1 x (as long as x = 0) and so we have y2 = 4 x. But now, what is y? y could be positive or negative – there will generally be two choices here, one positive and one negative. The appearance of two answers violates the uniqueness requirement in our

• utputs for a function.

Answer: NO, this is not a function. If x = 1 then we don’t know if y = 2 or y = −2.

Smith (SHSU) Elementary Functions 2013 17 / 27

Not a function

2 Does the equation xy2 = 4 define y as a function of x?

• Solution. If we attempt to solve for y, we multiply both sides of the

equation by 1 x (as long as x = 0) and so we have y2 = 4 x. But now, what is y? y could be positive or negative – there will generally be two choices here, one positive and one negative. The appearance of two answers violates the uniqueness requirement in our

• utputs for a function.

Answer: NO, this is not a function. If x = 1 then we don’t know if y = 2 or y = −2.

Smith (SHSU) Elementary Functions 2013 17 / 27

Not a function

2 Does the equation xy2 = 4 define y as a function of x?

• Solution. If we attempt to solve for y, we multiply both sides of the

equation by 1 x (as long as x = 0) and so we have y2 = 4 x. But now, what is y? y could be positive or negative – there will generally be two choices here, one positive and one negative. The appearance of two answers violates the uniqueness requirement in our

• utputs for a function.

Answer: NO, this is not a function. If x = 1 then we don’t know if y = 2 or y = −2.

Smith (SHSU) Elementary Functions 2013 17 / 27

Not a function

3 Does the equation x2y = 0 define y as a function of x? (Why/why not?)

• Solution. Although it might be tempting to solve for y, first notice that if x

is zero then y could be 0 or 1 or 2.71828 or anything! So the input x = 0 does not give a unique output. This is not a function. Answer: NO; if x = 0 then y could be anything. (This is different than problem 1. In problem 1, x = 0 is not a possible input in the equation. But here x = 0 is a possibility for a solution to the equation! So we have to worry about the input x.)

Smith (SHSU) Elementary Functions 2013 18 / 27

Not a function

3 Does the equation x2y = 0 define y as a function of x? (Why/why not?)

• Solution. Although it might be tempting to solve for y, first notice that if x

is zero then y could be 0 or 1 or 2.71828 or anything! So the input x = 0 does not give a unique output. This is not a function. Answer: NO; if x = 0 then y could be anything. (This is different than problem 1. In problem 1, x = 0 is not a possible input in the equation. But here x = 0 is a possibility for a solution to the equation! So we have to worry about the input x.)

Smith (SHSU) Elementary Functions 2013 18 / 27

Not a function

3 Does the equation x2y = 0 define y as a function of x? (Why/why not?)

• Solution. Although it might be tempting to solve for y, first notice that if x

is zero then y could be 0 or 1 or 2.71828 or anything! So the input x = 0 does not give a unique output. This is not a function. Answer: NO; if x = 0 then y could be anything. (This is different than problem 1. In problem 1, x = 0 is not a possible input in the equation. But here x = 0 is a possibility for a solution to the equation! So we have to worry about the input x.)

Smith (SHSU) Elementary Functions 2013 18 / 27

Not a function

3 Does the equation x2y = 0 define y as a function of x? (Why/why not?)

• Solution. Although it might be tempting to solve for y, first notice that if x

is zero then y could be 0 or 1 or 2.71828 or anything! So the input x = 0 does not give a unique output. This is not a function. Answer: NO; if x = 0 then y could be anything. (This is different than problem 1. In problem 1, x = 0 is not a possible input in the equation. But here x = 0 is a possibility for a solution to the equation! So we have to worry about the input x.)

Smith (SHSU) Elementary Functions 2013 18 / 27

Not a function

3 Does the equation x2y = 0 define y as a function of x? (Why/why not?)

• Solution. Although it might be tempting to solve for y, first notice that if x

is zero then y could be 0 or 1 or 2.71828 or anything! So the input x = 0 does not give a unique output. This is not a function. Answer: NO; if x = 0 then y could be anything. (This is different than problem 1. In problem 1, x = 0 is not a possible input in the equation. But here x = 0 is a possibility for a solution to the equation! So we have to worry about the input x.)

Smith (SHSU) Elementary Functions 2013 18 / 27

Not a function

3 Does the equation x2y = 0 define y as a function of x? (Why/why not?)

• Solution. Although it might be tempting to solve for y, first notice that if x

is zero then y could be 0 or 1 or 2.71828 or anything! So the input x = 0 does not give a unique output. This is not a function. Answer: NO; if x = 0 then y could be anything. (This is different than problem 1. In problem 1, x = 0 is not a possible input in the equation. But here x = 0 is a possibility for a solution to the equation! So we have to worry about the input x.)

Smith (SHSU) Elementary Functions 2013 18 / 27

Practicing function notation

Let us practice the function notation, f(x). A formula for f(x) tells us how the input x leads to the output f(x). For example, suppose f(x) = x2 − 9. Compute:

1 f(0), 2 f(1), 3 f(−1), 4 f(−5), 5 f(−x)

• Solutions. If f(x) = x2 − 9 then

1 f(0) = 02 − 9 = −9 . 2 f(1) = (1)2 − 9 = 1 − 9 = −8 . 3 f(−1) = (−1)2 − 9 = 1 − 9 = −8 , 4 f(−5) = (−5)2 − 9 = 25 − 9 = 16 . 5 f(−x) = (−x)2 − 9 = x2 − 9 .

Smith (SHSU) Elementary Functions 2013 19 / 27

Practicing function notation

Let us practice the function notation, f(x). A formula for f(x) tells us how the input x leads to the output f(x). For example, suppose f(x) = x2 − 9. Compute:

1 f(0), 2 f(1), 3 f(−1), 4 f(−5), 5 f(−x)

• Solutions. If f(x) = x2 − 9 then

1 f(0) = 02 − 9 = −9 . 2 f(1) = (1)2 − 9 = 1 − 9 = −8 . 3 f(−1) = (−1)2 − 9 = 1 − 9 = −8 , 4 f(−5) = (−5)2 − 9 = 25 − 9 = 16 . 5 f(−x) = (−x)2 − 9 = x2 − 9 .

Smith (SHSU) Elementary Functions 2013 19 / 27

Practicing function notation

Let us practice the function notation, f(x). A formula for f(x) tells us how the input x leads to the output f(x). For example, suppose f(x) = x2 − 9. Compute:

1 f(0), 2 f(1), 3 f(−1), 4 f(−5), 5 f(−x)

• Solutions. If f(x) = x2 − 9 then

1 f(0) = 02 − 9 = −9 . 2 f(1) = (1)2 − 9 = 1 − 9 = −8 . 3 f(−1) = (−1)2 − 9 = 1 − 9 = −8 , 4 f(−5) = (−5)2 − 9 = 25 − 9 = 16 . 5 f(−x) = (−x)2 − 9 = x2 − 9 .

Smith (SHSU) Elementary Functions 2013 19 / 27

Practicing function notation

Let us practice the function notation, f(x). A formula for f(x) tells us how the input x leads to the output f(x). For example, suppose f(x) = x2 − 9. Compute:

1 f(0), 2 f(1), 3 f(−1), 4 f(−5), 5 f(−x)

• Solutions. If f(x) = x2 − 9 then

1 f(0) = 02 − 9 = −9 . 2 f(1) = (1)2 − 9 = 1 − 9 = −8 . 3 f(−1) = (−1)2 − 9 = 1 − 9 = −8 , 4 f(−5) = (−5)2 − 9 = 25 − 9 = 16 . 5 f(−x) = (−x)2 − 9 = x2 − 9 .

Smith (SHSU) Elementary Functions 2013 19 / 27

Practicing function notation

Let us practice the function notation, f(x). A formula for f(x) tells us how the input x leads to the output f(x). For example, suppose f(x) = x2 − 9. Compute:

1 f(0), 2 f(1), 3 f(−1), 4 f(−5), 5 f(−x)

• Solutions. If f(x) = x2 − 9 then

1 f(0) = 02 − 9 = −9 . 2 f(1) = (1)2 − 9 = 1 − 9 = −8 . 3 f(−1) = (−1)2 − 9 = 1 − 9 = −8 , 4 f(−5) = (−5)2 − 9 = 25 − 9 = 16 . 5 f(−x) = (−x)2 − 9 = x2 − 9 .

Smith (SHSU) Elementary Functions 2013 19 / 27

Practicing function notation

Let us practice the function notation, f(x). A formula for f(x) tells us how the input x leads to the output f(x). For example, suppose f(x) = x2 − 9. Compute:

1 f(0), 2 f(1), 3 f(−1), 4 f(−5), 5 f(−x)

• Solutions. If f(x) = x2 − 9 then

1 f(0) = 02 − 9 = −9 . 2 f(1) = (1)2 − 9 = 1 − 9 = −8 . 3 f(−1) = (−1)2 − 9 = 1 − 9 = −8 , 4 f(−5) = (−5)2 − 9 = 25 − 9 = 16 . 5 f(−x) = (−x)2 − 9 = x2 − 9 .

Smith (SHSU) Elementary Functions 2013 19 / 27

Practicing function notation

Let us practice the function notation, f(x). A formula for f(x) tells us how the input x leads to the output f(x). For example, suppose f(x) = x2 − 9. Compute:

1 f(0), 2 f(1), 3 f(−1), 4 f(−5), 5 f(−x)

• Solutions. If f(x) = x2 − 9 then

1 f(0) = 02 − 9 = −9 . 2 f(1) = (1)2 − 9 = 1 − 9 = −8 . 3 f(−1) = (−1)2 − 9 = 1 − 9 = −8 , 4 f(−5) = (−5)2 − 9 = 25 − 9 = 16 . 5 f(−x) = (−x)2 − 9 = x2 − 9 .

Smith (SHSU) Elementary Functions 2013 19 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Practicing function notation

More examples. Let’s continue with the function f(x) = x2 − 9. Compute:

6 f(x + h), 7 f(√x), 8 f(2a + 1), 9 −f(x) + 2

• Solutions. If f(x) = x2 − 9 then

6 f(x + h) = (x + h)2 − 9 = (x2 + 2xh + h2) − 9 = x2 + 2xh + h2 − 9 , 7 f(√x) =(√x)2 − 9 = x − 9 , 8 f(2a + 1) = (2a + 1)2 − 9 = (4a2 + 4a + 1) − 9 = 4a2 + 4a − 8 , 9 −f(x) + 2 = −(x2 − 9) + 2 = −x2 + 11 .

Smith (SHSU) Elementary Functions 2013 20 / 27

Function notation

In the next presentation we find the domains of functions. (END)

Smith (SHSU) Elementary Functions 2013 21 / 27

Function notation

In the next presentation we find the domains of functions. (END)

Smith (SHSU) Elementary Functions 2013 21 / 27