MATH 12002 - CALCULUS I 2.1: Derivative and Slope Examples - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §2.1: Derivative and Slope Examples

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 8

Examples

Example

Find the equation of the tangent line to f (x) = 1

x at x = 4.

Solution

We need to determine the slope of the line and a point on the line. Since the line is tangent to the graph of f at x = 4, the line and the graph both pass through the point (4, f (4)) =

• 4, 1

4

• .

The slope of the line is f ′(4) = lim

h→0

f (4 + h) − f (4) h . [Continued →]

D.L. White (Kent State University) 2 / 8

Examples

Solution [continued]

Calculate the slope: f ′(4) = lim

h→0

f (4 + h) − f (4) h = lim

h→0 1 4+h − 1 4

h = lim

h→0 4 4(4+h) − 4+h (4+h)4

h = lim

h→0

• 4−(4+h)

4(4+h)

• h

= lim

h→0

4 − 4 − h 4(4 + h) · h = lim

h→0

−h 4(4 + h) · h = lim

h→0

−1 4(4 + h), since h = 0, = −1 4(4 + 0) = − 1 16. [Continued →]

D.L. White (Kent State University) 3 / 8

Examples

Solution [continued]

Hence the tangent line has slope − 1

16 and passes through the point (4, 1 4).

By the Point-Slope form of the equation of a line, the equation of the tangent line is y − 1 4 = − 1 16(x − 4), and simplifying yields y = − 1 16x + 4 16 + 1 4,

• r

y = − 1 16x + 1 2.

D.L. White (Kent State University) 4 / 8

Examples

Example

Let f (x) = x2 − 5x. Find f ′(2), f ′(3), and f ′(4).

Solution

Let’s first compute f ′(2): f ′(2) = lim

h→0

f (2 + h) − f (2) h = lim

h→0

[(2 + h)2 − 5(2 + h)] − [22 − 5(2)] h = lim

h→0

[4 + 4h + h2 − 5(2) − 5h] − [4 − 10] h = lim

h→0

4 + 4h + h2 − 10 − 5h − 4 + 10 h [Continued →]

D.L. White (Kent State University) 5 / 8

Examples

Solution [continued]

= lim

h→0

4 + 4h + h2 − 10 − 5h − 4 + 10 h = lim

h→0

−h + h2 h = lim

h→0

h(−1 + h) h = lim

h→0(−1 + h), since h = 0,

= −1 + 0 = −1. Hence f ′(2) = −1. We could find f ′(3) and f ′(4) using the same method, but it will be more efficient to find f ′(a) for any number a first. [Continued →]

D.L. White (Kent State University) 6 / 8

Examples

Solution [continued]

f ′(a) = lim

h→0

f (a + h) − f (a) h = lim

h→0

[(a + h)2 − 5(a + h)] − [a2 − 5a] h = lim

h→0

[a2 + 2ah + h2 − 5a − 5h] − [a2 − 5a] h = lim

h→0

a2 + 2ah + h2 − 5a − 5h − a2 + 5a h = lim

h→0

2ah + h2 − 5h h [Continued →]

D.L. White (Kent State University) 7 / 8

Examples

Solution [continued]

= lim

h→0

2ah + h2 − 5h h = lim

h→0

h(2a + h − 5) h = lim

h→0(2a + h − 5), since h = 0,

= 2a + 0 − 5 = 2a − 5. Therefore, f ′(a) = 2a − 5 for any number a. In particular, f ′(3) = 2(3) − 5 = 1 and f ′(4) = 2(4) − 5 = 3.

D.L. White (Kent State University) 8 / 8