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Find the local maximum and minimum values of the following: $2x^{3}+5x^{2}-4x $

Note: This is part 2nd of a 6 part question, split as 6 separate questions here.

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  • Second derivative test: Suppose $f$ is continous 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 $'c'$
$f(x)= 2x^3+5x^2-4x$
Step 1:
$f'(x)=6x^2+10 x-4$
At a turning point $f'(x)=0=>6x^2+10x-4=0$
$\therefore \;x=\large\frac{1}{3}$$,-2$
Step 2:
When $ x=\large\frac{1}{3}, f\bigg( \large\frac{1}{3}\bigg)$
$\qquad=2 \times \large\frac{1}{27}+\frac{5}{9}-\frac{4}{3}$
$\qquad= \large\frac{2+15-36}{27}$
$\qquad= \large\frac{-19}{27}$
When $ x=-2, f( 2)=-16+20+8$
The stationary points are $\bigg( \large\frac{1}{3},$$0\bigg),$$(-2,12)$
Step 3:
$f''(x)=12 x +10$
When $ x=\large\frac{1}{3}, f''\bigg( \large\frac{1}{3}\bigg)$$=14 > 0$=>This corresponds to a local minimum
When $ x=-2, f''(-2)$$=-24+10=-14 < 0$=>This corresponds to a local maximum
Step 4:
$\therefore$ there is a local minimum is at $\bigg(\large\frac{1}{3},\frac{-19}{27}\bigg)$
$\therefore$ there is a local maximum at $(-2,12)$


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