Finding the max value of a function in a given interval

Question

$f(x)=2x^5-5x^2$, interval [-2, 2]

How should I approach this, after I found the first derivative?

Answer 1

$f^{'}(x)=10x^4-10x=10x(x^3-1)$, so $f^{'}(x)=0$ only for $x=1$ and $x=0$. So you need to compute $f(0), f(1), f(-2), f(2)$. Max value from that set is the max value of $f(x)$.

The max value of a differentiable function on a closed interval can be attained in points where $f^{'}(x)=0$ or at the ends of the interval.

Answer 2

The maximum of a polynomial will occur at a critical point (i.e. where the derivative is zero), or at an endpoint of the interval. So if the derivative equals zero (in the given interval) at, say, $x_1$ and $x_2$, find $f(x_1), f(x_2), f(-2)$ and $f(2)$

Answer 3

You have arrived at $f'(x) = 10x^4 - 10x$, what's next?

One of the easy theorems of calculus say that a maximum of a function $f$ in some inverval can be attained either at the endpoint (in this case at -2, or 2), or at the point $x$ so that $f'(x) = 0$ (otherwise following the slope of the tangent, we would find a bigger value somewhere in the neighbourhood). $f'(x)$ is a polynomial, so it will have just a few roots that we need to check. Let's find them now, i.e., we want to solve $$ 10 x^4 - 10 x = 0. $$ That might not be easy in general, but we are lucky in this case since $x^4 - x = x(x^3-1)$, so the real roots are $x=0$ and $x = 1$.

Now, just check the four values $x=-2, 0, 1, 2$ which one gives the maximum. I'll leave that to you.