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Calculus AB: Applications of the Derivative


Problems for "Curve Sketching"

Optimization, page 2

page 1 of 3

Optimization of One Variable

For what value(s) of x will the function f (x) = x 3 -2x 2 - 5x be largest on the interval [- 10, 10] ?

This problem is asking us to find the absolute maximum on the interval. One way to do this is to graph the function and determine by inspection what the absolute maximum is. However, this is a very time-consuming process. A more streamlined approach is to identify all possible candidates for a local maximum and compare them to each other to see which is the largest.

The possible places for an absolute maximum to occur are at the local maxima or at the endpoints of the interval. The first step is to identify the critical points and classify them as local maxima, local minima, or neither.

f'(x) = x 2 - 4x - 5 = (x - 5)(x + 1),    

so f'(x) = 0 at x = 5 and x = - 1 .

Now, to classify these points, the second derivative test can be used.

f''(x) = 2x - 4 ;
f''(5) = 6 , which is greater than zero, so f (5) is a local minimum.
f''(- 1) = - 6 ,which is less than zero, so f (- 1) is a local maximum.

Finally, the actual values of the local maxima and the endpoints should be compared to each other to see which of these is the absolute maximum.

f (- 10) = - 283
f (- 1) = 2
f (10) = 83

So, the absolute maximum on the interval occurs when x = 10 .

Example: For what value(s) of x will the function f (x) = x 3 -2x 2 - 5x be largest for all x ?

This problem is identical to the one above, except that there are no endpoints to compare to the local maxima.

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