# `f(x) = sin(x), [(5pi)/6,(11pi)/6]` Find the absolute extrema of the function on the closed interval.

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### 1 Answer

Given the function: `f(x)=sin(x)` in the closed interval [5 pi/6 , 11 pi/6]

We have to find the absolute extrema of the function on the closed interval.

**Step 1: **Check whether f(x) is a continuous function.

Here f(x) is continuous since all sinusoidal function are continuous functions.

**Step 2: **Find the critical points.

To find the critical points we have to differentiate the function f(x)=sin(x) and equate it to zero.

So f'(x)=cos(x)=0`` implies x=`` ``n pi/2`` ``

`` ``But we need only the critical points that fall inside the closed interval

[5 pi/6 , 11 pi/6]

Therefore the critical points are: pi and 3pi/2 .``

**Step 3: **Now we have to find the values of the function at the two critical points and the end points of the interval.

`f(pi)=sin(pi)=0`

`f((3pi)/2)=sin((3pi)/2)=-1`

`f((5pi)/6)=sin((5pi)/6)=1/2`

`f((11pi)/6)=sin((11pi)/6)=-1/2`

Absolute extrema are the largest and the smallest the function will ever be. So we can see that the absolute maximum of f(x) is `1/2`and it occurs at x= 5 pi/6(end point) and the absolute minimum of f(x) is -1 and it occurs at x= 3 pi/2 (critical point).