Question

Find the absolute extrema of f(x) = x^2/3(x^2 − 1) on the interval [−1, 8].

please help

Answer 1

By long division,\[\frac{x^2}{3(x^2-1)}=\frac{1}{3}+\frac{1}{3(x^2-1)}\]which essentially reduces the problem to finding the extrema of \(\dfrac{1}{x^2-1}\). This in turn can be written as
\[\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)\]To summarize,
\[\frac{x^2}{3(x^2-1)}=\frac{1}{3}+\frac{1}{6}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)\]All this really did was simplify the function so you can avoid using the quotient rule (not that it's terribly hard to go through with this function).

Compute your derivative:
\[\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{3}+\frac{1}{6}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)\right]=\frac{1}{6}\left(\frac{1}{(x+1)^2}-\frac{1}{(x-1)^2}\right)\]Find your critical points - where is the derivative zero? undefined?
\[\frac{1}{6}\left(\frac{1}{(x+1)^2}-\frac{1}{(x-1)^2}\right)=0~~\iff~~(x+1)^2=(x-1)^2~~\implies~~\cdots\]Finding out where it's undefined should be obvious.

Once you find your critical points \(x\), consider some test points between them and check the value of the derivative over each interval. Also check the values of the function at the endpoints at \([-1,8]\).

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