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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.

Toolbox:
• 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 $f'(x)=0=>6x^2+10x-4=0$
$3x^2+5x-2=0$
$3x^2+6x-x-2=0$
$3x(x+2)-1(x+2)=0$
$(3x-1)(x+2)=0$
$\therefore \;x=\large\frac{1}{3}$$,-2 Step 2: When x=\large\frac{1}{3}, f\bigg( \large\frac{1}{3}\bigg) \qquad=2 \times \large\frac{1}{27}+\frac{5}{9}-\frac{4}{3} \qquad= \large\frac{2+15-36}{27} \qquad= \large\frac{-19}{27} When x=-2, f( 2)=-16+20+8 \qquad=12 The stationary points are \bigg( \large\frac{1}{3},$$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 $(-2,12)$

edited Aug 1, 2013 by meena.p