Section 2.7 Derivatives and Rates of Change

Section 2.7 Derivatives and Rates of Change

EX.8

Find an equation of the tangent line to the curve at the given point.

y = 2x + 1

x + 2 , (1, 1) (sol)

m = lim

x→1

2x+1 x+2−1

x−1 = lim

x→1 x−1

(x−1)(x+2) = ¹₃ tangent line : y − 1 = ¹₃(x − 1)

EX.10

(a) Find the slope of the tangent to the curve y = ^√1_x at the point where x = a.

(b) Find equations of the tangent lines at the points (1, 1) and (4,¹₂).

(c) Graph the curve and both tangents on a common screen.

(sol)

(a) m = lim

x→a

√1 x−^√1

a

x−a = lim

x→a

−√ x+√

√ a ax(√

x−√ a)(√

x+√

a) = _2a^−1√_a (b)







a = 1 ⇒ m = −¹₂ a = 4 ⇒ m = −₁₆¹

⇒ tangent line :







y − 1 = −¹₂(x − 1) at (1, 1) y − ¹₂ = −₁₆¹(x − 4) at (4,¹₂) (c)

EX.20

If the tangent line to y = f (x) at (4, 3) passes through the point (0, 2), find f (4) and f⁰(4).

(sol)

f (4) = 3

m = ³⁻²₄₋₀ = ¹₄ = f⁰(4)

1

EX.22

Sketch the graph of a function g for which

g(0) = g(2) = g(4) = 0, g⁰(1) = g⁰(3) = 0, g⁰(0) = g⁰(4) = 1 g⁰(2) = −1, lim_x→∞g(x) = ∞, and lim_x→−∞g(x) = −∞

(sol)

EX.31

Find f⁰(a)

f (x) =√ 1 − 2x (sol)

f⁰(a) = lim

x→a

f (x) − f (a) x − a

= lim

x→a

√1 − 2x −√ 1 − 2a

x − a = lim

x→a

(1 − 2x) − (1 − 2a) (x − a)(√

1 − 2x +√

1 − 2a)

= lim

x→a

−2 (√

1 − 2x +√

1 − 2a) = −1

√1 − 2a

EX.35

Each limit represents the derivative of some function f at some number a. State such an f and a in each case.

x→5lim

2^x− 32 x − 5 (sol)

f (x) = 2^x, a = 5

2

EX.36

Each limit represents the derivative of some function f at some number a. State such an f and a in each case.

x→lim^π₄

tan x − 1 x − ^π₄ (sol)

f (x) = tan x, a = ^π₄

3