Question

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

Answer 1

Absolute max: `x = pi/8`
Absolute min. is at the endpoints: `x = 0, x = pi/4`

Explanation:

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

`y' = (cosu) u' - (sinu) u' = 2cos2x - 2sin2x`

Find critical numbers by setting `y' = 0` and factor:
`2(cos2x-sin2x) = 0`

When does `cosu = sinu`? when `u = 45^@ = pi/4`
so `x = u/2 = pi/8`

Find the 2nd derivative: `y''= -4sin2x-4cos2x`

Check to see if you have a max at `pi/8` using the 2nd derivative test:

`y''(pi/8) ~~-5.66 < 0`, therefore `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]}

Answer 2

`0 and sqrt2`. See the illustrative Socratic graph.

Explanation:

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

Use `| sin (theta) | in [0, 1]`.

`|f|=|sin2x+cos2x|`

`sqrt2|sin2x cos(pi/4)+cosx sin(pi/4)|`

`=sqrt2|sin(2x+pi/4)| in [0, sqrt 2]`.