# How do you find the intervals in which the function f(x) = sin^4x + cos^4x is increasing and decreasing in [0, pi/2]?

Jun 29, 2017

#### Explanation:

Investigate the sign of $f ' \left(x\right)$ on the interval $\left[0 , \frac{\pi}{2}\right]$

$f ' \left(x\right) = 4 {\sin}^{3} x \cos x - 4 {\cos}^{3} x \sin x$

$f '$ is defined for all $x$ and we need to find its zeros.

$f ' \left(x\right) = 4 \sin x \cos x \left({\sin}^{2} x - {\cos}^{2} x\right) = 0$

A product is zero if and only if one of its factors is zero.

Obviously, $4$ cannot be $0$

$\sin x = 0$ at $x = 0$ $\text{ }$ (Remember we have a restricted domain.)

$\cos x = 0$ at $x = \frac{\pi}{2}$

${\sin}^{2} x - {\cos}^{2} x = 0$ at $x = \frac{\pi}{4}$ $\text{ }$ (See note below.)

In $\left[0 , \frac{\pi}{4}\right]$ test $f ' \left(\frac{\pi}{6}\right)$ to see that $f '$ is negative, so $f$ is decreasing.

In $\left[\frac{\pi}{4} , \frac{\pi}{2}\right]$ test $f ' \left(\frac{\pi}{3}\right)$ to see that $f '$ is positive, so $f$ is increasing.

Note

Here are three ways to solve ${\sin}^{2} x - {\cos}^{2} x = 0$

Use the double angle formula for cosine

${\sin}^{2} x - {\cos}^{2} x = - \left({\cos}^{2} x - {\sin}^{2} x\right) = - \cos \left(2 x\right) = 0$ at $2 x = \frac{\pi}{2} + 2 \pi k$ and at $x = \frac{3 \pi}{2} + 2 \pi k$. The only solution with $x$ in $\left[0 , \frac{\pi}{2}\right]$ is $x = \frac{\pi}{4}$

Factor first

$\left(\sin x + \cos x\right) \left(\sin x - \cos x\right) = 0$

$\sin x + \cos x = 0$ has no solutions in the first quadrant.

$\sin x - \cos x = 0$ iff $\sin x = \cos x$, so $x = \frac{\pi}{4}$.

Divide first

${\sin}^{2} x = {\cos}^{2} x$ iff ${\sin}^{2} \frac{x}{{\cos}^{2} x} = 1$ which requires $\tan x = \pm 1$. In the first quadrant the only solution is $x = \frac{\pi}{4}$.