Question

What are the points of inflection of #f(x)=8x^2 + sin(2x-pi) `on` x in [0, 2pi]#?

Answer 1

None. No points of inflection.

Explanation:

Points of inflection are slope extrema or `f'(x)` extrema. To find a maximum or minimum, take the second derivative and set equal to zero, to find where change in slope is zero.

`f(x)=8x^2+sin(2x-pi)`
`f'(x)=16x + 2cos(2x-pi)`
`f''(x)=16 - 4sin(2x-pi)`

Now set `f''(x)=0`:

`0=16-4sin(2x-pi)`
`-16=-4sin(2x-pi)`
`4=sin(2x-pi)`
`sin^-1(4)=2x-pi`
`->`NO SOLUTION

There are no (real, I dunno about complex) solutions as sin(x) cannot be greater than 1.

Therefore there are no points of inflection.