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

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Answer 1 · 2017-02-18T07:56:57

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

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 · 2017-02-18T09:04:56

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

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

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

$|f|=|sin2x+cos2x|$

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

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

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