# The maximum value of the function f(x)=(3cosx-2)^2-2 ?

May 22, 2018

Maximum value = 23

#### Explanation:

$f \left(x\right) = {\left[3 \cos \left(x\right) - 2\right]}^{2} - 2$

$f ' \left(x\right) = 2 \sin \left(x\right) \left[3 \cos \left(x\right) - 2\right] \cdot \left[- 3 \sin \left(x\right)\right]$
$f ' \left(x\right) = - 6 \sin \left(x\right) \left[3 \cos \left(x\right) - 2\right] = 0$
$\sin \left(x\right) \left[3 \cos \left(x\right) - 2\right] = 0$

$\sin \left(x\right) = 0$
$x = 2 k \pi \pm \pi$
$x = \pi$
#f(pi)=[-3-2]^2-2=25-2=23

$3 \cos \left(x\right) - 2 = 0$
$\cos \left(x\right) = \frac{2}{3}$
$x = 2 k \pi \pm {\cos}^{-} 1 \left(\frac{2}{3}\right)$
$f \left(2 k \pi \pm {\cos}^{-} 1 \left(\frac{2}{3}\right)\right) = - 2$

Obviously the minimum value of the function is -2 and the maximum value of the function is 23

May 22, 2018

f(x)max = -1

#### Explanation:

The unique variable here is cos x that reaches max when cos x = 1
The maximum value is:
$f \left(x\right) = {\left(3 - 2\right)}^{2} - 2 = 1 - 2 = - 1$