MATH 12002 - CALCULUS I 5.3: The Natural Exponential Function - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §5.3: The Natural Exponential Function

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 7

Definition & Basic Properties

Recall f (x) = ln x is one-to-one with domain (0, ∞) and range (−∞, ∞). Therefore, ln x has an inverse, which for now we call will f −1(x) = exp x, with domain (−∞, ∞) and range (0, ∞). By the cancellation property of inverses, ln(exp x) = x for all x. Recall also that we defined the number e with the property that ln e = 1. By the Laws of Logarithms, if x is any real number, then ln(ex) = x ln e = x · 1 = x. Hence for any real number x, we have ln(exp x) = ln(ex). Since ln x is one-to-one, this implies exp x = ex for all x, and so ex is precisely the inverse of our natural logarithm function. In particular, ln(ex) = x for all x eln x = x for all x > 0. The function y = ex is called the natural exponential function.

D.L. White (Kent State University) 2 / 7

Definition & Basic Properties

As the inverse of y = ln x, the graph of the natural exponential function is the reflection in the line y = x of the graph of y = ln x:

§5.3 FIGURE 1

Thus the graph of y = ex is as in the figure below:

§5.3 FIGURE 2 D.L. White (Kent State University) 3 / 7

Definition & Basic Properties

A number of properties of ex follow from properties of its inverse ln x: ln x = y ⇐ ⇒ ey = x. y = ex has domain (−∞, ∞) and range (0, ∞). That is, ex is defined for all x, but ex is never 0 or negative. e0 = 1 and e1 = e, so the points (0, 1) and (1, e) are on the graph. y = ln x is differentiable everywhere and its derivative ( 1

x ) is never 0,

hence y = ex is differentiable (so also continuous) everywhere. y = ex is increasing and concave up everywhere. Since y = ln x has vertical asymptote x = 0, y = ex has horizontal asymptote y = 0, i.e., lim

x→−∞ ex = 0.

Since lim

x→+∞ ln x = +∞, we also have

lim

x→+∞ ex = +∞.

D.L. White (Kent State University) 4 / 7

Solving Equations

Inverse operations are often used in solving equations: Subtract 5 to solve x + 5 = 3 (inverse of addition). Divide by 5 to solve 5x = 3 (inverse of multiplication). Take the 5th root to solve x5 = 3 (inverse of 5th power). We can use ex and ln x to solve logarithmic and exponential equations. Examples:

1 Solve for x: ln(5x + 2) = 8.

Since ln x = y ⇔ ey = x, we can translate this equation to the exponential equation e8 = 5x + 2, 5x = e8 − 2, x = (e8 − 2)/5 ≈ 595.79.

D.L. White (Kent State University) 5 / 7

Solving Equations

2 Solve for x: e2x−3 = 4.

To get at the variable in the exponent, we apply the inverse function y = ln x to both sides of the equation: ln e2x−3 = ln 4 2x − 3 = ln 4 x = (3 + ln 4)/2 ≈ 2.19.

3 Solve for x: 2x−3 = 10.

Although ln x is not the inverse of 2x, we can still apply ln x to both sides of the equation and apply Laws of Logarithms: ln 2x−3 = ln 10 (x − 3) ln 2 = ln 10 x − 3 = ln 10/ ln 2 x = 3 + ln 10 ln 2 ≈ 6.32.

D.L. White (Kent State University) 6 / 7

Evaluating Limits

We noted previously that lim

x→+∞ ex = +∞ and

lim

x→−∞ ex = 0.

Examples:

1 Evaluate the limit

lim

x→+∞ e5−x2.

As x → +∞, 5 − x2 → −∞, so lim

x→+∞ e5−x2 =

lim

z→−∞ ez = 0.

2 Evaluate the limit

lim

x→+∞

2ex − 5e−x 3ex + 7e−x . As x → +∞, ex → +∞ and e−x → 0. Similar to what we did with rational functions, we can divide numerator and denominator by the term that approaches ∞: lim

x→+∞

2ex − 5e−x 3ex + 7e−x = lim

x→+∞

(2ex − 5e−x) 1

ex

(3ex + 7e−x) 1

ex

= lim

x→+∞

2 − 5e−2x 3 + 7e−2x = 2 − 0 3 + 0 = 2 3.

D.L. White (Kent State University) 7 / 7