# How do you find the global extreme values for f(t) = 2cost + sin2t on [0,pi/2]?

Aug 27, 2015

Use the closed interval method.

#### Explanation:

Find any critical numbers in $\left[0 , \frac{\pi}{2}\right]$,

then find the value of $f$ at $0$, the critical numbers in $\left[0 , \frac{\pi}{2}\right]$, and at $\frac{\pi}{2}$.

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

$f ' \left(t\right) = - 2 \sin t + 2 \cos 2 t$

 = -2(sint - (1-2sin^2t)

$= - 2 \left(2 {\sin}^{2} t + \sin t - 1\right)$

$= - 2 \left(2 \sin t - 1\right) \left(\sin t + 1\right)$.

So $f ' \left(t\right)$ is never undefined and is $0$ where:

$\sin t = \frac{1}{2}$ $\text{ }$ or $\text{ }$ $\sin t = - 1$.

The only solution in the interval $\left[0 , \frac{\pi}{2}\right]$ is $t = \frac{\pi}{6}$.

$f \left(0\right) = 2 \cos \left(0\right) + \sin 2 \left(0\right) = 2$ .

$f \left(\frac{\pi}{6}\right) = 2 \cos \left(\frac{\pi}{6}\right) + \sin \left(\frac{\pi}{3}\right) = \frac{2 \sqrt{3}}{2} + \sqrt{3} = \frac{3 \sqrt{3}}{2}$.

$f \left(\frac{\pi}{2}\right) = 2 \cos \left(\frac{\pi}{2}\right) + \sin \pi = 0$ .

The minimum is $0$ (and it occurs at $\frac{\pi}{2}$).

The maximum is $\frac{3 \sqrt{3}}{2}$ (and it occurs at $\frac{\pi}{6}$).

Note:
We can see that $\frac{3 \sqrt{3}}{2} > 2$ by using the fact that, for numbers greater than 1, the square of the greater number is greater.

The square of $\frac{3 \sqrt{3}}{2}$ is $\frac{9 \cdot 3}{4} = \frac{27}{4}$

while the square of $2$ is $4 = \frac{16}{4}$.

So $\frac{3 \sqrt{3}}{2} > 2$.