Figuring out when f(x)=sin(x^2) is increasing and decreasing.

Silvina2b 2022-08-11 Answered
Figuring out when f ( x ) = sin ( x 2 ) is increasing and decreasing
Regarding the function f ( x ) = sin ( x 2 ), I'm supposed to figure out when it is increasing/decreasing.
So far, I've found the derivative to be f ( x ) = 2 x cos ( x 2 ).
So long as I can solve the inequality cos ( x 2 ) > 0. I can figure the rest out, but this is where I'm stuck.
I've narrowed it down to π 2 < x 2 < π 2 meaning x 2 < π 2 ± n 2 π ,     n N .
Then I get x < ± π 2 ± n 2 π .
Am I on the right track here? I can't seem to find a path from here.
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Answers (1)

Addison Herman
Answered 2022-08-12 Author has 15 answers
Step 1
You're on the right track.
The "points that matter" in partitioning the increasing/decreasing intervals are the points for which f ( x ) = 0. This happens when either 2 x = 0 or cos ( x 2 ) = 0.
The function is increasing when f ( x ) > 0, and decreasing when f ( x ) < 0.
Consider each part of the derivative separately.
The linear part of the derivative, 2x, is positive when x > 0, and negative when x < 0.
Step 2
The cosine function part of the derivative, cos ( x 2 ), is positive when its argument is on one the open intervals
( 2 π n π / 2 , 2 π n + π / 2 ) , n Z ,
and negative when it's on one the open intervals ( 2 π n + π / 2 , 2 π n + 3 π / 2 ) , n Z .
The next steps would involve looking at when the product of these two parts changes sign. Those points would define your increasing/decreasing intervals.

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