How do I find the maximum and minimum values of the function $f (x) = x - 2 sin (x)$ $f(x) = x - 2 sin (x)$ on the interval $[- \frac{π}{4}, \frac{π}{2}]$ $[-pi/4, pi/2]$ ?

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How do I find the maximum and minimum values of the function $f (x) = x - 2 sin (x)$ $f(x) = x - 2 sin (x)$ on the interval $[- \frac{π}{4}, \frac{π}{2}]$ $[-pi/4, pi/2]$ ?

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Answer 1 · 2015-01-30T00:36:10

You can derive your function and set your derivative equal to zero. The value(s) of $x$ $x$ you'll find will be the points of maxima or minima.

In your case you have:

$f'(x)=1-2cos(x)$

Setting $f'(x)=0$ gives:

$1-2cos(x)=0$
When $x=pi/3$

You can now analyze when the derivative is bigger than zero:

$1-2cos(x)>0$
i.e. when $x>pi/3$
enter image source here
Your value of $x=pi/3$ represent a minimum for your function, and in your interval you get the graph:

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How do I find the maximum and minimum values of the function $f (x) = x - 2 sin (x)$ $f(x) = x - 2 sin (x)$ on the interval $[- \frac{π}{4}, \frac{π}{2}]$ $[-pi/4, pi/2]$ ?

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How do I find the maximum and minimum values of the function $f (x) = x - 2 sin (x)$ $f(x) = x - 2 sin (x)$ on the interval $[- \frac{π}{4}, \frac{π}{2}]$ $[-pi/4, pi/2]$ ?