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Find the local maximum and minimum values of the following: $x^{3}-x$

Note: This is part 1st of a 6 part question, split as 6 separate questions here.

1 Answer

  • Second derivative test: Suppose $f$ is continuous on an open interval that contains
  • (i)If $f'(c)=0\; and\; f''(c) >0$ then $f$ has a local minimum at $c$.
  • (ii)If $f'(c)=0\;and \; f''(c)<0$ then $f$ has a local maximum at $'c'$
Step 1:
At a turning point $f'(x)=0,$ gives $3x^2-1=0$
=> $x=\pm \large\frac{1}{\sqrt 3}$
Step 2:
When $ x=\large\frac{1}{\sqrt 3}, f\bigg( \large\frac{1}{\sqrt 3}\bigg)=\frac{1}{3 \sqrt 3}-\frac{1}{\sqrt 3}$
$\qquad=\large\frac{-2}{3 \sqrt 3}$
When $ x=\large\frac{-1}{\sqrt 3}, f\bigg( \large\frac{-1}{\sqrt 3}\bigg)=\frac{1}{3 \sqrt 3}+\frac{1}{\sqrt 3}$
$\qquad=\large\frac{2}{3 \sqrt 3}$
The stationary points are $\bigg(\large\frac{1}{\sqrt 3},\frac{-2}{3 \sqrt 3}\bigg),\bigg(\large\frac{-1}{\sqrt 3},\frac{2}{3 \sqrt 3}\bigg)$
Step 3:
When $ x=\large\frac{1}{\sqrt 3}, f''\bigg( \large\frac{1}{\sqrt 3}\bigg)=\frac{6}{\sqrt 3}$$ > 0$=>This corresponds to a local minimum
When $ x=\large\frac{-1}{\sqrt 3}, f''\bigg( \large\frac{-1}{\sqrt 3}\bigg)=\frac{-6}{\sqrt 3}$$ < 0$=>This corresponds to a local maximum
Step 4:
$\therefore$ there is a local minimum is at $\bigg(\large\frac{1}{\sqrt 3},\frac{-2}{3 \sqrt 3}\bigg)$
$\therefore$ there is a local maximum at $\bigg(\large\frac{-1}{\sqrt 3},\frac{2}{3 \sqrt 3}\bigg)$


answered Aug 1, 2013 by meena.p
edited Aug 1, 2013 by meena.p