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

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

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`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...

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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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