Optimization practice question 5
Find the absolute maximum value of on the closed interval .
On a closed interval, the absolute extremes occur at a critical point, at a singular point (where f′ does not exist) or at an endpoint. All of them must be checked.
factors as .
Theory UBC Math 100 §8.2CLP-1 §3.5Key results
Worked solution
Strategy
Use the closed-interval method: find the critical and singular points inside the interval, evaluate the function at those points and at both endpoints, and compare the list.
Working
Differentiate and factor. Both critical points lie inside [0, 5], and a polynomial has no singular points.
Evaluate at the endpoints. Skipping these is the usual way to lose this question.
Evaluate at the critical points.
The largest value is at the endpoint x = 5, well above the interior local maximum f(1) = 6.
Answer
The answer is 22, at the endpoint x = 5.
Check
f(x) − 22 = (x − 5)(x² − x + 4), and x² − x + 4 is always positive, so f(x) ≤ 22 for every x up to 5.
Where this goes wrong
Checking only the critical points and answering with the interior local maximum f(1) = 6, or answering f(3) = 2 by picking the wrong critical point. On a closed interval the endpoints are candidates too, and here x = 5 gives the largest value.
The idea behind it
The optimization lesson, a figure to move, in a few steps
Leans on
Critical points, from Differentiation rules
Try a similar question
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