What are the extrema of $f(x)=2x^3 + 5x^2 - 4x - 3$ ?

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Answer 1 · 2016-12-10T12:37:54

$x_1=-2$ is a maximum

$x_2=1/3$ is a minimum.

First we identify the critical points by equating the first derivative to zero:

$f'(x) = 6x^2 +10x -4 = 0$

giving us:

$x= frac (-5 +- sqrt (25+24)) 6 = (-5 +- 7)/6$

$x_1= -2$ and $x_2=1/3$

Now we study the sign of the second derivative around the critical points:

$f''(x) = 12x+10$

so that:

$f''(-2) < 0$ that is $x_1=-2$ is a maximum

$f''(1/3) > 0$ that is $x_2=1/3$ is a minimum.

graph{2x^3+5x^2-4x-3 [-10, 10, -10, 10]}

What are the extrema of f(x)=2x^3 + 5x^2 - 4x - 3 f(x)=2x^3 + 5x^2 - 4x - 3 ?