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

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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)`.**