# Thread: Having trouble with graph of trig. function

1. ## Having trouble with graph of trig. function

I'm doing a graph sketching of $f(x)=sin^4(x) + cos^4(x)$
And I'm having trouble with two things:

1)
I get $f''(x)=-4sin^4(x) -4cos^4(x) + 24sin^2(x)cos^2(x)$
Then, I checked on WolframAlpha and gives me an alternate form :
$f''(x)=-4cos(4x)$ which is equivalent to what i got. But I have no idea how to go from the first form to the latter.

2)
When taking limits of f(x) as x-> infinity and -infinity, I understand:
$lim (sin(x))$as x->+-inifinty ; i get y=1 and y=-1 as the horizontal asymptotes.
Also:
$lim (sin(x))^4$ as x->+-infinity; i get y=1 and y=0 as the horizontal asymptotes.

Then for lim of cos^4(x) as x->+-infinity I get the same thing, but when those two are added, the graph is between the lines y=.5 and y=1

Any help would be greatly appreciated.

2. Originally Posted by Arturo_026
I'm doing a graph sketching of $f(x)=sin^4(x) + cos^4(x)$
And I'm having trouble with two things:

1)
I get $f''(x)=-4sin^4(x) -4cos^4(x) + 24sin^2(x)cos^2(x)$
Then, I checked on WolframAlpha and gives me an alternate form :
$f''(x)=-4cos(4x)$ which is equivalent to what i got. But I have no idea how to go from the first form to the latter.

[snip]
$-4[\sin^4(x) + \cos^4(x) - 6 \sin^2(x) \cos^2(x)]$

$= -4 [(\sin^2 (x) + \cos^2 (x))^2 - 8\sin^2(x) \cos^2(x)]$

$= -4 [1 - 4\sin^2(2x)]$

which won't quite give -4 cos(4x). Is there a typo somewhere?

3. Originally Posted by mr fantastic
$-4[\sin^4(x) + \cos^4(x) - 6 \sin^2(x) \cos^2(x)]$

$= -4 [(\sin^2 (x) + \cos^2 (x))^2 - 8\sin^2(x) \cos^2(x)]$

$= -4 [1 - 4\sin^2(2x)]$

which won't quite give -4 cos(4x). Is there a typo somewhere?
I apologize for jumping a while ago.

after: $= -4 [(\sin^2 (x) + \cos^2 (x))^2 - 8\sin^2(x) \cos^2(x)]$
shouldn't it be:

$= -4[1-2sin^2(2x)]$
$=-4cos(4x)$

4. Originally Posted by Arturo_026
I apologize for jumping a while ago.

after: $= -4 [(\sin^2 (x) + \cos^2 (x))^2 - 8\sin^2(x) \cos^2(x)]$
shouldn't it be:

$= -4[1-2sin^2(2x)]$
$=-4cos(4x)$
Right you are. I was in a hurry. I'm glad to see you got the idea of what to do.