You can
put this solution on YOUR website!f(x) = cos(x)(cos(x) + 1)
f'(x) = (cos(x))'(cos(x) + 1) + (cos(x))(cos(x) + 1)'
f'(x) = -sin(x)(cos(x) + 1) + cos(x)(-sin(x))
f'(x) = -sin(x)cos(x) - sin(x) - cos(x)sin(x)
f'(x) = -sin(x)(2cos(x) + 1)
Slope of tangent (defined by f'(x)) is zero at max and min values.
0 = -sin(x)(2cos(x) + 1)
sin(x) = 0 when x = 0,
, 2
, 3
...
cos(x) = -1/2 when x = 2
/3, 4
/3 ...
No worries, functions of cosine and sine occillate here.
f(x) = cos(x)^2 + cos(x)
f(0) = 1^2 + 1 = 2
f(
) = (-1)^2 - 1 = 0
f(2
) = 1^2 + 1 = 2
f(2
/3) = (-0.5)^2 - 0.5 = -0.25
f(4
/3) = (-0.5)^2 - 0.5 = -0.25
Maximum for f(x) is 2.
~ Proof ~
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