Question

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?

Answer 1

Critical point: number `c` in the domain of `f` with `f'(c)=0` or #f'(c) does not exist.

Finding Critical Points
`f(x)=x^3-15x^2+4`

`f'(x)=3x^2-30x`

`f'(x)` does not exist -- no such `x`

`f'(x)=3x^2 - 30x=0`
`3x(x-10)=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''(x)=6x-30`

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

At the critical point `10`, we have `f''(10)=6(10)-30 > 0`
The second derivative test (for local extrema) tells us that, `f(10)` 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]} #