Chapter 4 | Applications of Derivatives
371
a. The derivative f ′( x ) = x 2 −5 x +4 is defined for all real numbers x . Therefore, we only need to find the values for x where f ′( x ) =0. Since f ′( x ) = x 2 −5 x +4= ( x −4)( x −1), the critical points are x =1 and x =4. From the graph of f in Figure 4.16 , we see that f has a local maximum at x =1 and a local minimum at x =4.
Figure 4.16 This function has a local maximum and a local minimum.
b. Using the chain rule, we see the derivative is
2
2
f ′( x ) =3 ⎛
⎞ ⎠
(2 x ) =6 x ⎛
⎞ ⎠
⎝ x 2 −1
⎝ x 2 −1
.
Therefore, f has critical points when x =0 andwhen x 2 −1=0. We conclude that the critical points are x =0, ±1. From the graph of f in Figure 4.17 , we see that f has a local (and absolute) minimum at x =0, but does not have a local extremum at x =1 or x =−1.
Figure 4.17 This function has three critical points: x =0, x =1, and x =−1. The function has a local (and absolute) minimum at x =0, but does not have extrema at the other two critical points.
c. By the chain rule, we see that the derivative is
⎛ ⎝ 1+ x 2 4 ⎞
⎠ −4 x (2 x )
x 2
f ′( x ) =
= 4−4
2 .
2
⎛ ⎝ 1+ x 2
⎞ ⎠
⎛ ⎝ 1+ x 2
⎞ ⎠
The derivative is defined everywhere. Therefore, we only need to find values for x where f ′( x ) =0. Solving f ′( x ) =0, we see that 4−4 x 2 =0, which implies x =±1. Therefore, the critical points are x =±1. From the graph of f in Figure 4.18 , we see that f has an absolute maximum at x =1
Made with FlippingBook - professional solution for displaying marketing and sales documents online