Maxima and minima second derivative

Question

Let $f(x) = 2x^{3} - x^{4} - 10$ over $[-1,1]$

I found second derivative as $12x - 12x^2$

After trying values I found that $f(x)$ gives minimum value at $x=-1$

If I substitute $x=-1$ in second derivative, then I get $-24<0$ so it should be Maxima right Should we check second derivative only for critical points ? If yes, then why?

Answer 1

$$f(x) = 2x^3 - x^4 - 10 $$ It is defined, continuous, and differentiable at every point in $[-1,1]$

$$f'(x)=6x^2-4x^3$$ $$f''(x)=12x-12x^2$$

Let $f'(x)=0$, we find $x=0, \frac{3}{2}$

For $x \in [-1,1]$, we exclude $x=\frac{3}{2}$.

$f''(0)=0$, thus we cannot directly tell if it is max or min. or neither.

Thus we now complete value at $0$ and at boundaries:

$$f(0)=-10, \, f(1)=-9, \,f(-1)=-13$$

Thus max=$-9$, and min=$-13$

Or even better, we could see that: $$f'(x)=6x^2-4x^3 \ge 0 \text{ over }[-1,1]$$

Thus it is non-decreasing over $[-1,1]$, and we get the result immediately

Answer 2

To answer your question: $f''(x_0)<0$ implies $f(x)$ is concave at the point $x_0$. If $x_0$ is a critical point ($f'(x_0)=0$), then it is a relative maximum. In your case $-1$ is not a critical point (it is a border point). Your problem is called a constrained optimization problem. According to the Extreme value theorem, the function is continuous in a closed interval, so it must have max and min. Now its extreme (max/min) values will be at either critical or border points, which is what you calculated. In general, the relative max could be not max of a constrained optimization problem. For example: find max of $f(x)=x^3-3x$ in $[-3,3]$.