Question

Find the local maximum or minimum of any of the function f(x) =sin4x+cos4x

Answer 1

Given the function $f(x)=\sin 4x+\cos 4x$. This function can be rewritten as

$f(x)=\sqrt{2}(\sin 4x \frac{1}{\sqrt{2}}+\cos 4x \frac{1}{\sqrt{2}}) \\ f(x)=\sqrt{2}(\sin 4x \cos (45^o)+\cos 4x \sin (45^o)) \\ f(x)=\sqrt{2} \sin (4x+45^o)$

Now the maximum value of $f(x)=\sqrt{2} \sin (4x+45^o)$ is $f(x)=\sqrt{2}$ and minimum value is $f(x)=-\sqrt{2}$.

Maximum value occurs when $\sin (4x+45^o)=1$ and minimum value occurs when $\sin (4x+45^o)=-1$