How do you find the local maximum and minimum values of $f(x)=2x^3 + 5x^2 - 4x - 3$ ?

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Answer 1 · 2017-06-10T14:39:18

$f(-2)=9$ is local Maxima, $f(1/3)=-100/27$ is local

Minima.

We know that, for local Extreme values, $f'(x)=0.$

Also, $f''(x)<0$ for Maxima, and, $f''(x) >0$ for Minima.

$f(x)=2x^3+5x^2-4x-3$

$rArr f'(x)=6x^2+10x-4, &, f''(x)=12x+10.$

$f'(x)=0 rArr 2(3x^2+5x-2)=0.$

$rArr 2(x+2)(3x-1)=0.$

$rArr x=-2, x=1/3.$

Now, $f''(-2)=-24+10=-14 < 0.$

$:. f$ has a local maxima at $x=-2,$ &, it is,

$f(-2)=-16+20+8-3=9.$

Also, $f''(1/3)=4+10=14 > 0.$

$:. f$ has a local minima at $x=1/3,$ which is,

$f(1/3)=2/27+5/9-4/3-3=-100/27.$

Thus, $f(-2)=9$ is local Maxima, $f(1/3)=-100/27$ is local

Minima.

Enjoy Maths.!

How do you find the local maximum and minimum values of f(x)=2x^3 + 5x^2 - 4x - 3f(x)=2x^3 + 5x^2 - 4x - 3?