Find the absolute maximum and minimum values of $f(x) = e^{\cos x}$ over the interval $[-\pi,2\pi]$

Question

I know the derivative is

$$-e^{\cos x}\sin x$$

and that I have to set the derivative equal to $0$.

Beyond that I can't figure out how to isolate for $x$.

Answer 1

$f(x)=e^{\cos(x)}$

1)$ \exp $ is a strictly increasing function , I.e.

$x_1 \lt x_2$ implies $e^{x_1} \lt e^{x_2}.$

2) $\cos(x)$: $x \in [-π,2π]$ , I.e. $\cos(x)$ attains its minimum $= -1$, and its maximum $=1$ in this interval,
since $[0,2π] \subset [-π,2π]$, I.e. comprises a full cycle of $\cos(x)$.

Hence :

$\min (f(x))= e^{-1}$ , $x\in [-π,2π].$ $\max (f(x))=e^{+1}$, $x \in [-π,2π].$

At which points are these values attained?

Answer 2

Guide:

Note that exponential function is an increasing function.

If $a \leq \cos x \leq b$, then $\exp(a) \leq \exp(\cos(x)) \leq \exp(b)$.

Answer 3

It's also always a good idea to check the endpoints of a closed interval because extreme values can (and will) occur there for a continuous function on a compact domain. The key insight is that setting a derivative equal to zero only captures local information about the function, and, even if the function is differentiable, it's not sufficient for the function to have a local maximum or minimum. For example $f(x) = x^3$ has $f'(0)=0$, and the function is increasing everywhere.

Also, here's an important technical point. The derivative of a function is only defined on open sets, so the derivative does not exist at the boundary of a closed interval if that is the function's domain.

You know that the maximum of $cos(x)$ on $[-\pi, 2 \pi]$ occurs at $x = 0$ and $2\pi$. You know that its minimum occurs at $-\pi$ and $\pi$. Since $e^x$ is strictly increasing, the hint above tells you everything you need to know.