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

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Answer 1 · 2018-05-22T09:55:25

Maximum value = 23

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

$f'(x)=2sin(x)[3cos(x)-2] * [-3sin(x)]$
$f'(x)= -6sin(x)[3cos(x)-2]=0$
$sin(x)[3cos(x)-2]=0$

$sin(x)=0$
$x=2kpi+-pi$
$x=pi$
#f(pi)=[-3-2]^2-2=25-2=23

$3cos(x)-2=0$
$cos(x)=2/3$
$x=2kpi+-cos^-1(2/3)$
$f(2kpi+-cos^-1(2/3))=-2$

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

Answer 2 · 2018-05-22T14:18:13

f(x)max = -1

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