Question

f(n)=sin4x+cos4x local max and min?

Answer 1

work with
\[\sqrt{2}\sin(x+\frac{\pi}{4})\]

Answer 2

oops it is 4x, ok
\[\sqrt{2}\sin(4x+\frac{\pi}{4})\]

Answer 3

so max is
\[\sqrt{2}\] and min is
\[-\sqrt{2}\]

Answer 4

The answer appears to be local min pi/4, 3pi/4, 5pi/4, 7pi/4 and local max; pi/2 and 3pi/2

Answer 5

i will respectfully disagree

Answer 6

first of all max in an output, not an input

Answer 7

shoot, im an idiot. I meant \[f(n)=\sin^4x+\cos^4x \]

Answer 8

oooooooooooooooooooh

Answer 9

then i still respectfully disagree as for the reason above,
max is an output not an input
it will have a max when \(x=\frac{\pi}{4}\) and at that point you will get 1, so the max is one

Answer 10

sorry what you said, max at \(x=\frac{\pi}{2}\) but the max is 1

Answer 11

but my calculus textbook! it says otherwise. Can you show me the method?