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Find the local maximum and minimum values of the following: $(x^{2}-1)^{3}$

Note: This is part 4th 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 $'c'$
$f(x)=(x^2-1)^3$
Step 1:
$f'(x)=3 (x^2-1)^2.2x=6x(x^2-1)^2$
At a turning point $f'(x)=0=>6x(x^2-1)^2=0$
$=>x=0,-3,+1$
Step 2:
When $x=0,f(0)=1$
When $x=-1,f(-1)=0$
When $x=1,f(1)=0$
The stationary points are at $(0,-1),(-1,0),(1,0)$
Step 3:
$f''(x)=6(x^2-1)^2+6x.2(x^2-1).2x$
$\qquad \quad=6(x^2-1)^2+24x^2(x^2-1)$
When $x=0,f''(0)=6-24=-18$=> This corresponds to a local maximum
When $x=-1,f''(x)=0$=> The second derivative test fails
When $x=1,f''(x)=0$=> The second derivative test fails
$\therefore \; f(x)$ has a local maximum at $(0,-18)$

 

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