How do you find the maximum of #f(x) = 2sin(x^2)#?

Question

Paisley Johnson · Accepted Answer

The maximum value of #f(x)=2sin(x^2)# is #2#. This is an example of a kind of problem sometimes asked in a calculus class.  It is a kind of a trap question.  You do not need calculus to answer this question. The maximum value of the sine function is #1#. This happens when #x^2=pi/2+2 pi k# for any integer #k#.  I also know that #x^2= pi/2# when #x=sqrt(pi/2)#. So the maximum value of this function is #2#. If you need to know where else the maximum occurs, then you need to solve: #x^2=pi/2+2 pi k=pi/2 (1+4k)# for any integer #k#. The solutions are  #x=+-sqrt(pi/2)sqrt (1+4k)# which might be easier to read written:  #x=+-sqrt (1+4k)sqrt(pi/2)# for any integer k.

Ezekiel Alexander · Answer

undefined To find the maximum of $ f(x) = 2\sin(x^2) $, you first need to find the critical points by setting the derivative equal to zero and solving for $ x $. Then, you evaluate the second derivative at those critical points to determine if they correspond to maxima or minima. Finally, you compare the function values at the critical points and endpoints to find the maximum value.

First, find the derivative of $ f(x) $ with respect to $ x $:

$ f'(x) = 4x\cos(x^2) $

Set $ f'(x) = 0 $ to find critical points:

$ 4x\cos(x^2) = 0 $

This equation is satisfied when $ x = 0 $ or $ \cos(x^2) = 0 $.

$ x = 0 $ is a critical point.

To find where $ \cos(x^2) = 0 $, solve $ x^2 = \frac{\pi}{2} + n\pi $ for $ x $, where $ n $ is an integer:

$ x = \sqrt{\frac{\pi}{2} + n\pi} $ or $ x = -\sqrt{\frac{\pi}{2} + n\pi} $

These are additional critical points.

Next, evaluate the second derivative $ f''(x) $:

$ f''(x) = 4\cos(x^2) - 8x^2\sin(x^2) $

Now, evaluate $ f''(x) $ at the critical points:

$ f''(0) = 4 $ (since $ \cos(0) = 1 $)

$ f''\left(\sqrt{\frac{\pi}{2} + n\pi}\right) = -4\sqrt{\frac{\pi}{2} + n\pi} $ for all integer values of $ n $

$ f''\left(-\sqrt{\frac{\pi}{2} + n\pi}\right) = 4\sqrt{\frac{\pi}{2} + n\pi} $ for all integer values of $ n $

Since $ f''(0) > 0 $, $ x = 0 $ corresponds to a local minimum. For the other critical points, $ f''(x) $ is negative, indicating they are maxima.

Now, evaluate $ f(x) $ at the critical points and endpoints to find the maximum value:

$ f(0) = 0 $

$ f\left(\sqrt{\frac{\pi}{2} + n\pi}\right) = 2\sin\left(\frac{\pi}{2} + n\pi\right) = 2 $ for all integer values of $ n $

$ f\left(-\sqrt{\frac{\pi}{2} + n\pi}\right) = -2 $ for all integer values of $ n $

So, the maximum value of $ f(x) $ is $ 2 $.