184 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(1.9) 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(1.9)≈12(1.9)− 16 = 6. 8
f(2.01)≈12(2.01)− 16 = 8. 12 .(See Figure 9.2-2.)
y
f (x) = x^3
(2, 8)
0
x
1.9 2 2.01
Tangent line
Not to Scale
y = 12 x – 16
Figure 9.2-2
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 isy− 6 =−
1
2
(x−2) ory=−
1
2
x+7;
(d) thus,f(2.1)≈−