Question

what are the x-coordinates for the maximum points in the function f(x) = 4 cos(2x- pi) from x = 0 to x= 2 pi?

Answer 1

Answer:

$\frac{\pi }{2}$ and $\frac{3\pi }{2}$

Step-by-step explanation:

To find the max points we need to take the derivative of the function and then find the critical values.

First we take the derivative:

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

Now we need to find when f'(x)=0 to find the critical values.

$0=-8sin(2x-\pi )\\0=sin(2x-\pi )\\sin^{-1}0=2x-\pi \\0=2x-\pi \\\pi =2x\\\frac{\pi }{2} =x$

The critical values will be

$\frac{\pi }{2} n$ for any integer n

between 0 and 2 pi, the critical values will be

$0, \frac{\pi }{2} ,\pi ,\frac{3\pi }{2},2\pi$

We can determine if these are minimums or maximums by using the second derivative test.

So we need to take the second derivative;

$f'(x)=-8sin(2x-\pi )\\f''(x) = -8cos(2x-\pi )(2)\\f''(x)=-16cos(2x-\pi)$

We need to see if the second derivative is positive or negative to determine if it is a max or min.

$f''(0) = 16\\f''(\frac{\pi}{2})=-16\\f''(\pi )=16\\f''(\frac{2\pi}{3}) = -16$

Since the second derivative is negative at

$\frac{\pi }{2}$ and $\frac{3\pi }{2}$

we know both of those are the x-values of maximums.