`f(x) = x - ln(x), [(1/2), 2]` Find the absolute maximum and minimum values of f on the given interval

Textbook Question

Chapter 4, 4.1 - Problem 60 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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This function is continuous on the given interval and is differentiable inside it. So it reaches minimum and maximum either at endpoint or where f'(x)=0.

f(1/2) = 1/2-ln(1/2)=1/2 + ln(2).

f(2)=2-ln(2).

f'(x)=1-1/x. It is zero at x=1. f(1)=1.

Which number is max and which is min?

1/2 + ln(2) > 1 because ln(2)>1/2 because 2>sqrt(e) because 4>e.

2-ln(2)>1 because 1>ln(2) because e>2.

And 1/2+ln(2) < 2-ln(2) because ln(2) < 1-1/4 because` 2 lt e^(1-1/4) approx 2.117` (don't know how check this without calculator).

The answer: minimum is 1 and maximum is 2-ln2.

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