Question

PLEASE HELP!!!!!!!!!!
On what intervals within [0, 2π) is f increasing? (Enter your answer using interval notation. If an answer does not exist, enter DNE.)
y=(-√2) cos(πx/2-π/4)

Answer 1

lets to this the boneheaded way.
cosine is increasing on \((-\pi,\pi)\) so \(-\cos(x)\) is increasing on \((0,\pi)\)

Answer 2

set
\[\frac{\pi x}{2}-\frac{\pi}{4}=0\] and solve for x

Answer 3

you get
\[\frac{\pi x}{2}=\frac{\pi}{4}\]
\[x=\frac{1}{2}\]

Answer 4

then set
\[\frac{\pi x}{2}-\frac{\pi}{4}=\pi\]
\[\frac{\pi x}{2}=\frac{5\pi}{4}\]
\[x=\frac{5\pi}{2}\]

Answer 5

sorry last one should just be
\[x=\frac{5}{2}\]

Answer 6

so increasing on \((\frac{1}{2},\frac{5}{2})\)

Answer 7

it says it's wrong((

Answer 8

y=(-√2) cos(πx/2-π/4)
y'=+pie/root(2) [sin(πx/2-π/4)]
@x=1/2 => y'=0
therefore , f increases from [0,1/2]

Answer 9

either wrong(

Answer 10

looks right to me
http://www.wolframalpha.com/input/?i=y%3D%28-%E2%88%9A2%29+cos%28%CF%80x%2F2-%CF%80%2F4%29
i am sticking with my answer