5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 15:14

212 STEP 4. Review the Knowledge You Need to Score High

Example 1

Write an equation of the tangent line tof(x)=x^3 at (2, 8). Use the tangent line to find the

approximate values off(1.90) andf(2.01).

Differentiatef(x):f′(x)= 3 x^2 ; f′(2)=3(2)^2 =12. Since f is differentiable atx=2, an

equation of the tangent atx=2 is:

y=f(2)+f′(2)(x−2)

y=(2)^3 +12(x−2)= 8 + 12 x− 24 = 12 x− 16

f(1.90)≈12(1.90)− 16 = 6. 80

f(2.01)≈12(2.01)− 16 = 8. 12 .(See Figure 10.2-2.)

y

f (x) = x^3

(2, 8)

0

x

1.9 2 2.01

Tangent line

Not to scale

y = 12 x – 16

Figure 10.2-2

Example 2

Iff is a differentiable function and f(2)=6 and f′(2)=−

1

2

, find the approximate value

off(2.1).

Using tangent line approximation, you have

(a) f(2)= 6 ⇒the point of tangency is (2, 6);

(b) f′(2)=−

1

2

⇒the slope of the tangent atx=2ism=−

1

2

;

(c) the equation of the tangent line isy− 6 =−

1

2

(x−2) ory=−

1

2

x+7;

(d) thus,f(2.1)≈−

1

2

(2.1)+ 7 ≈ 5 .95.