# Stewart Calculus Book Answers

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**Calculus homework help!?**

This is a problem from Calculus Early Transcendentals, James Stewart, 6 Edition. The section is 4.1 and problem 59. You are trying to find absolute extrema and the problem is f(x)=xe^((x^2)/8), and the answer in the back of the book is 2 (the interval is bounded from -1 to 4) and 1. and i cannot figure out where the 1 came from.

f(x) = xe^(x²/8)

Extrema occur where f'(x) = 0, so find all roots of f'(x) between -1 and 4.

f'(x) = ¼x²e^(x²/8) + e^(x²/8)

f'(x) = (¼x² + 1)(e^(x²/8))

e^(x²/8) is an exponential, so it is never equal to zero. So f'(x) is 0 when (¼x² + 1) = 0.

(¼x² + 1) = 0

¼x² = -1

x² = -4

x = 2i, x = -2i

There are no real zeros to f'(x), so f(x) has no natural extrema. The extrema are only present because the interval is restricted. So one extrema will be at f(-1) and the other at f(4).

f(-1) = -e^(1/8) = -1.13 —–> Minimum

f(4) = 4e² = 29.6 ——> Maximum

I don’t know how your book got that answer, but nothing of interest happens at y = 2, or y = 1. f(x) itself has a root and switches concavity at x = 0, but that’s about the only thing going on with this curve.

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