# How do you locate the critical points of the function f(x) = x^3 - 15x^2 + 4 and use the Second Derivative Test to determine whether they correspond to local maxima, local minima, or neither?

Mar 26, 2015

Critical point: number $c$ in the domain of $f$ with $f ' \left(c\right) = 0$ or f'(c) does not exist.

Finding Critical Points
$f \left(x\right) = {x}^{3} - 15 {x}^{2} + 4$

$f ' \left(x\right) = 3 {x}^{2} - 30 x$

$f ' \left(x\right)$ does not exist -- no such $x$

$f ' \left(x\right) = 3 {x}^{2} - 30 x = 0$
$3 x \left(x - 10\right) = 0$ at $x = 0$, $x = 10$

Both $0$ and $10$ are in the domain of $f$, so they are both crical points for $f$

Testing Critical Points
The second derivative of $f$:

$f ' ' \left(x\right) = 6 x - 30$

At the critical point $0$, we have $f ' ' \left(0\right) = - 30 < 0$
The second derivative test (for local extrema) tells us that, $f \left(0\right)$ is a local maximum. There is a local maximum of $4$ at $0$..

At the critical point $10$, we have $f ' ' \left(10\right) = 6 \left(10\right) - 30 > 0$
The second derivative test (for local extrema) tells us that, $f \left(10\right)$ is a local minimum. There is a local minimum of $- 496$ at $10$..

And here's the graph (you'll have to zoom to see details):

graph{y=x^3-15x^2+4 [-16, 41.74, -19.96, 8.92]}