`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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calendarEducator since 2015
write762 answers
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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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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