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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.
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).
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