# What are the absolute extrema of f(x)=2cosx+sinx in[0,pi/2]?

May 1, 2017

Absolute max is at $f \left(.4636\right) \approx 2.2361$

Absolute min is at $f \left(\frac{\pi}{2}\right) = 1$

#### Explanation:

$f \left(x\right) = 2 \cos x + \sin x$

Find $f ' \left(x\right)$ by differentiating $f \left(x\right)$
$f ' \left(x\right) = - 2 \sin x + \cos x$

Find any relative extrema by setting $f ' \left(x\right)$ equal to $0$:
$0 = - 2 \sin x + \cos x$

$2 \sin x = \cos x$

On the given interval, the only place that $f ' \left(x\right)$ changes sign (using a calculator) is at

$x = .4636476$

Now test the $x$ values by plugging them into $f \left(x\right)$, and don't forget to include the bounds $x = 0$ and $x = \frac{\pi}{2}$

$f \left(0\right) = 2$

$\textcolor{b l u e}{f \left(.4636\right) \approx 2.236068}$

$\textcolor{red}{f \left(\frac{\pi}{2}\right) = 1}$

Therefore, the absolute maximum of $f \left(x\right)$ for $x \in \left[0 , \frac{\pi}{2}\right]$ is at $\textcolor{b l u e}{f \left(.4636\right) \approx 2.2361}$, and the absolute minimum of $f \left(x\right)$ on the interval is at $\textcolor{red}{f \left(\frac{\pi}{2}\right) = 1}$