# Let f(x)= cos x + 2 cos^2 x defined for π/2 <x< π. Determine the local extrema and inflection points.

Asked on by cspanutius

lfryerda | High School Teacher | (Level 2) Educator

Posted on

To find the local extrema, we need to take the derivative and set it to zero.

f'(x)=-sin x-4sin x cos x

=-sin x(1+cos x)

When we set it to zero this means that the extrema are located where

1+4 cos x=0

cos x = -1/4

which is approximately at x=1.823 radians and at

sin x=0

which is at x=pi

This point is just outside the boundary of the interval, and so it won't be included. The only extrema is at the point (1.823, -1/8).

To find the inflection point, take the second derivative and set it to zero.

f''(x)=-cos x - 4 cos^2 x + 4 sin^2 x

=-cos x - 4 cos^2 x + 4 - 4 cos^2x

=-(8cos^2x +cos x - 4)

Setting the second derivative to zero gives a quadratic equation in cos x, which in this case is solved with the quadratic formula to give

cos x = {-1+-\sqrt{65}}/{16}

In solving this equation by taking cosine inverse of both sides, the positive root gives a solution outside the interval, and the negative root gives a solution of

x=2.1729

This means that the inflection point in the interval is (2.1729,-173.3).

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