You need to solve the equation `f'(x)=0` to find the critical values of function such that:

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

`f'(t) = 0 =gt -2sin t + 2 cos 2t = 0`

You need to substitute `1 - 2sin^2 t` for `cos 2t` in equation `-2sin t + 2 cos 2t = 0` such that:

`-2sin t + 2(1 - 2sin^2 t) =` 0

`-2 sin t + 2 - 4 sin^2 t = ` 0

You need to divide by -2 such that:

`2sin^2 t + sin t - 1 = 0`

You should come up with the substitution `sin t = y` such that:

`2y^2 + y - 1 = 0`

You need to solve for y the equation `2y^2 + y - 1 = 0` such that:

`y_(1,2) = (-1+-sqrt(1 + 8))/4`

`y_1 = (-1+3)/4 =gt y_1 = 1/2`

`y_2 = (-1-3)/4 =gt y_1 = -1`

You need to solve for t the equations `sint = 1/2` and `sint = -1` such that:

`sint = 1/2`

You need to remember that the values of sine function rest positive in 1 and 2 quadrants such that:

`t= pi/6 ; t= pi-pi/6 = 5pi/6`

`sint = -1 =gtt = 3pi/2`

The critical values of function gives the extrema of function.

You need to select a value for t smaller than `pi/6` such that:

`f(pi/8) = 2 cos pi/8 + sin pi/4 = 2*sqrt((1+cos(pi/4))/2) + sqrt2/2gt0`

You need to select a value for `t in (pi/6 , 5pi/6)` such that:

`f(pi/4) = 2cos pi/4 + sin pi/2 = 2sqrt2/2 + 1 gt 0`

You need to select a value for t larger than `5pi/6` such that:

`f(pi) = 2 cos pi + sin 2pi = -2 + 0 = -2 lt 0`

Notice that the function reaches its maximum at `t = 5pi/6` .

You need to select a value for t larger than `3pi/2 ` such that:

`f(11pi/6) = 2 cos (11pi/6) + sin 11pi/3 = 2 cos pi/6 - sin pi/3` (reduce to the first quadrant)

`f(11pi/6) = sqrt3 - sqrt3/2 = sqrt3/2 gt 0`

Notice that the function reaches its minimum at `t = 3pi/2` .

**Hence, evaluating the absolute minimum and maximum values of function yields that the function reaches its maximum at `t = 5pi/6` and the function reaches its minimum at `t = 3pi/2` .**

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