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Chapter 4 | Applications of Derivatives
Figure 4.19 This function has both an absolute maximum and an absolute minimum.
b. Step 1. Evaluate f at the endpoints x =0 and x =2.
f (0) =0and f (2) =4−3 4 3 ≈ −0.762
Step 2. The derivative of f is given by
4/3 −2 x 1/3
= 2 x
f ′( x ) =2 x − 2 x 1/3
for x ≠0. The derivative is zero when 2 x 4/3 −2=0, which implies x =±1. The derivative is undefined at x =0. Therefore, the critical points of f are x =0, 1, −1. The point x =0 is an endpoint, so we already evaluated f (0) in step 1. The point x =−1 is not in the interval of interest, so we need only evaluate f (1). We find that f (1) =−2.
Step 3. We compare the values found in steps 1 and 2, in the following table. x f ( x ) Conclusion
0 0
Absolute maximum
1 −2
Absolute minimum
2 −0.762
We conclude that the absolute maximum of f over the interval [0, 2] is zero, and it occurs at x =0. The absolute minimum is −2, and it occurs at x =1 as shown in the following graph.
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