# Find the exact global maximum and minimum values ( y values) of the function f(t)= x+...

## Question:

Find the exact global maximum and minimum values ({eq}y {/eq} values) of the function {eq}f(t)= x+ \sin(2x) {/eq} on the interval {eq}[0, \pi] {/eq}

## Maximum and Minimum:

Only if the function is continuous can we guarantee that the absolute ends of a function are reached over a closed and bounded range:

this will be either at a critical point or at the ends of the interval

First, we need to calculate the critical points that belongs to the interval:

{eq}f(x) = x + \sin (2x)\quad ,\left[ {0,\pi } \right]\\ \\ f'(x) = 1 + \cos (2x) \cdot 2 = 0\\ 2\cos \left( {2x} \right) = - 1\\ \cos (2x) = - \frac{1}{2}\\ \left\{ \begin{array}{l} 2x = \frac{{2\pi }}{3} \to x = \frac{\pi }{3}\\ 2x = \frac{{4\pi }}{3} \to x = \frac{{2\pi }}{3} \end{array} \right. {/eq}

As we can see, we have two critical points. To these two critical points we need to add to additional values, the ends of the interval:

{eq}\left\{ \begin{array}{l} f(0) = 0 + \sin (0) = 0\\ f(\pi /3) = \pi /3 + \sin (2\pi /3) \approx 1.91\\ f(2\pi /3) = 2\pi /3 + \sin (4\pi /3) \approx 1.23\\ f(\pi ) = \pi + \sin (2\pi ) \approx 3.14 \end{array} \right. {/eq}

So, the absolute minimum of the function is 0 and the absolute maximum of the function is 3.14.