How do you find all the extrema for $f (x) = \frac{x}{x^{2} + x + 1}$ $f(x) = x/(x^2+x+1)$ on [-2,2]?

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Answer 1 · 2015-09-10T13:43:24

Relative Max: $\frac{1}{3}$ $1/3$ (at $1$ $1$ ), Relative min $- 1$ $-1$ (at $- 1$ $-1$ )
Absolute max and min are the same as the relative.

$f (x) = \frac{x}{x^{2} + x + 1}$ $f(x) = x/(x^2+x+1)$ on [-2,2]

$f'(x) = (1-x^2)/(x^2+x+1)^2$

$f'(x)$ is never undefined and is $0$ at $x = +-1$

On $[-2,-1)$ , we have $f'(x) <0$
on $(-1,1)$ , we have $f'(x) > 0$

So $f(-1) = -1$ is a relative minumum.

On $(-1,1)$ , we have $f'(x) >0$
on $(1,2]$ , we have $f'(x) < 0$

So $f(1) = 1/3$ is a relative minumum.

$f(-2) = -2/3 > -1$ and $f(2) = 2/7 < 1/3$ so the relative extrema are also absolute on the interval.

How do you find all the extrema for f(x) = x/(x^2+x+1)f(x)=xx2+x+1f(x) = x/(x^2+x+1) on [-2,2]?