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Find the local maximum and minimum values of the following: $x^{4}-6x^{2}$

Note: This is part 3rd of a 6 part question, split as 6 separate questions here.

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Toolbox:
  • 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 $c$
$f(x)=x^4-6x^2$
Step 1:
$f'(x)=4x^3-12x=4x(x^2-3)$
At a turning point $f'(x)=0$ => $4x(x^2-3)$
$\therefore x=0, \pm \sqrt 3$
Step 2:
When $x=0\; f(0)=0$
When $x=\sqrt 3\; f(\sqrt 3)=9-18=-9$
When $x=-\sqrt 3\; f(-\sqrt 3)=-9$
The stationary points are $(0,0),(\sqrt 3, -9),(-\sqrt 3, -9)$
Step 3:
$f'(x)=12 x^2-12$
When $x=0, f''(0)=-12 < 0$ This corresponds to a local maximum
When $x=\sqrt 3, f''(\sqrt 3)=36 -12=24 > 0$ This corresponds to a local minimum
When $x=-\sqrt 3, f''(-\sqrt 3)=24 > 0$ This corresponds to a local minimum
Step 4:
$\therefore $ there is a local maximum at (0,0) and local minimum at $(\pm \sqrt 3,-9)$

 

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