# `f(t) = 2cos(t) + sin(2t), [0, pi/2]` Find the absolute maximum and minimum values of f on the given interval

gsarora17 | Certified Educator

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The function has absolute minimum=0 at x=`pi/2`

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## Related Questions

gsarora17 | Certified Educator

calendarEducator since 2015

starTop subjects are Math, Science, and Business

`f(t)=2cos(t)+sin(2t)`

differentiating,

`f'(t)=-2sint+2cos(2t)`

Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

Now...

(The entire section contains 2 answers and 181 words.)

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scisser | Student

First, find the derivative and set it equal to 0.

`f’(t)=-2sint+2cos2t=0`

`-sint+cos2t=0`

Using the double angle formula, `cos(2t) = 1 − 2 sin2(t)` , we then find

`−2 sin2(t) − sin(t) + 1 = 0`

Factor

`−(2 sin(t) − 1)(sin(t) + 1) = 0`

Therefore, `sint=1/2 or sint=-1` Since the interval is between `0 and pi/2` , then for the first value, `t=pi/6` . No value of t inside the interval is equal to -1.

Therefore,

only critical point in the interval `[0, pi/2]` is at `t = pi/6` . The boundary points of the interval are `0 and pi/2` . Since `f (0) = 2, f (pi/2) = 0, and f (pi/6) = (3/2)sqrt3` , we can conclude that

absolute min occurs at `x = pi/2` with value 0

and

absolute max occurs at `x = pi/4` with value (3/2)sqrt3

Therefore,

only critical point in the interval [0, pi/2] is at t = pi/6. The boundary points of the interval are 0 and pi/2.  Since f (0) = 2, f (pi/2) = 0, and f (pi/6) = (3/2)sqrt3, we can conclude that

absolute min occurs at x = pi/2 with value 0

and

absolute max  occurs at  x = pi/4   with value (3/2)sqrt3

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