`f(t) = 2cos(t) + sin(2t), [0, pi/2]` Find the absolute maximum and minimum values of f on the given interval
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 to find the critical numbers, solve for t for f'(t)=0.
`sin(t)=-1 , 1/2`
`sin(t)=1/2=> t=pi/6+2n*pi , (5pi)/6+2n*pi`
So , t=pi/6 in the interval (0,pi/2).
Evaluating the function at the critical point pi/6 and at the end points of the interval (o,pi/2),
So , the function has absolute maximum =`(3sqrt(3))/2` , at t=`pi/6`
The function has absolute minimum=0 at x=`pi/2`
First, find the derivative and set it equal to 0.
Using the double angle formula, `cos(2t) = 1 − 2 sin2(t)` , we then find
`−2 sin2(t) − sin(t) + 1 = 0`
`−(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.
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
absolute max occurs at `x = pi/4` with value (3/2)sqrt3
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
absolute max occurs at x = pi/4 with value (3/2)sqrt3