How do you find the critical points of y=x-2sinx on the interval [0, pi/2]?

Mar 2, 2018

There is one critical point of $y = x - 2 \sin x$ on the interval $\left[0 , \frac{\pi}{2}\right]$ at $\left(\frac{\pi}{3} , \frac{\pi}{3} - \sqrt{3}\right) = \left(1.047 , - .685\right)$

Explanation:

The critical points of a function occur where the derivative of the function is zero on the specified interval.

Find the derivative of $y$:
$y = x - 2 \sin x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 - 2 \cos x$

Set the derivative, $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to zero:
$0 = 1 - 2 \cos x$

$2 \cos x = 1$

$\cos x = \frac{1}{2}$

Within the interval from $\left[0 , \frac{\pi}{2}\right]$, picture the unit circle, and where is $\text{cosine of x}$ equal to $\frac{1}{2}$ $x = \frac{\pi}{3}$

$x \approx 1.047$

Therefore, there is only one critical point of $y$ on the provided interval, and this is where $x = \frac{\pi}{3}$.

The $y$ value at this point is:

$y = x - 2 \sin x$

$y = \frac{\pi}{3} - 2 \sin \left(\frac{\pi}{3}\right)$

$y = \frac{\pi}{3} - 2 \left(\frac{\sqrt{3}}{2}\right)$

$y = \frac{\pi}{3} - \sqrt{3}$

$y \approx - .685$

We can check our answer graphically. The slope changes from negative to positive at approximately 1.05.
graph{x-2sinx [-10, 10, -5, 5]}