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

1 Answer
Apr 30, 2018

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.