What are the absolute extrema of f(x)=sin2x + cos2x in[0,pi/4]?

Feb 18, 2017

Absolute max: $x = \frac{\pi}{8}$
Absolute min. is at the endpoints: $x = 0 , x = \frac{\pi}{4}$

Explanation:

Find the first derivative using the chain rule:
Let u = 2x; u' = 2, so $y = \sin u + \cos u$

$y ' = \left(\cos u\right) u ' - \left(\sin u\right) u ' = 2 \cos 2 x - 2 \sin 2 x$

Find critical numbers by setting $y ' = 0$ and factor:
$2 \left(\cos 2 x - \sin 2 x\right) = 0$

When does $\cos u = \sin u$? when $u = {45}^{\circ} = \frac{\pi}{4}$
so $x = \frac{u}{2} = \frac{\pi}{8}$

Find the 2nd derivative: $y ' ' = - 4 \sin 2 x - 4 \cos 2 x$

Check to see if you have a max at $\frac{\pi}{8}$ using the 2nd derivative test:

$y ' ' \left(\frac{\pi}{8}\right) \approx - 5.66 < 0$, therefore $\frac{\pi}{8}$ is the absolute max in the interval.

Check the endpoints:
y(0) = 1; y(pi/4) = 1 minimum values

From the graph:
graph{sin(2x) + cos(2x) [-.1, .78539816, -.5, 1.54]}

Feb 18, 2017

$0 \mathmr{and} \sqrt{2}$. See the illustrative Socratic graph.

Explanation:

graph{(|sin(2x)+cos(2x)|-y)y(y-sqrt2)=0 [-4, 4, 2, 2]}

Use $| \sin \left(\theta\right) | \in \left[0 , 1\right]$.

$| f | = | \sin 2 x + \cos 2 x |$

$\sqrt{2} | \sin 2 x \cos \left(\frac{\pi}{4}\right) + \cos x \sin \left(\frac{\pi}{4}\right) |$

$= \sqrt{2} | \sin \left(2 x + \frac{\pi}{4}\right) | \in \left[0 , \sqrt{2}\right]$.