How do you find the critical points for $f (x) = x^{3} - 15 x^{2} + 4$ $f(x) = x^3 - 15x^2 + 4$ and the local max and min?

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Answer 1 · 2016-12-21T15:15:01

The critical points are: $x_{1} = 0$ $x_1=0$ and $x_{2} = 10$ $x_2=10$ .
$x_{1}$ $x_1$ is a maximum, while $x_{2}$ $x_2$ is a minimum.

To identify the critical points, we have to solve the equation:

$f'(x)=0$

that is:

$3x^2-30x=0$

$3x(x-10) =0$

so that the critical points for $f(x)$ are $x_1=0$ and $x_2=10$ .
We can now calculate the second derivative:

$f''(x) = 6x-30$

and note that around $x_1$ $f''(x)$ is negative, therefore $x_1$ is a maximum, while around $x_2$ it is positive, therefore $x_2$ is a minimum.

graph{x^3-15x^2+4 [-80, 80, -640, 640]}

How do you find the critical points for f(x) = x^3 - 15x^2 + 4f(x)=x3−15x2+4f(x) = x^3 - 15x^2 + 4 and the local max and min?