Question

What are the points of inflection, if any, of `f(x) = 5xcos^2x − 10sinx^2` on `x in [0,2pi]`?

Answer 1

Find `f''(x)` and determine where it changes signs.

`f'(x)=5cos^2x-10xcosxsinx-20xcos(x^2)`

`f''(x)=-20cosxsinx+10xsin^2x-10xcos^2x-20cos(x^2)+40x^2sin(x^2)`

Graph of `f''(x)`:

graph{-20cos(x)sin(x)+10x(sinx)^2-10x(cosx)^2-20cos(x^2)+40x^2sin(x^2) [-0, 6.283, -1800, 1800]}

There are a slew of times when `f''(x)` switches from positive to negative on the interval, all of which are points of inflection.

They are:

• `x=0.886`
• `x=1.854`
• `x=2.521`
• `x=3.066`
• `x=3.559`
• `x=3.996`
• `x=4.341`
• `x=4.698`
• `x=5.011`
• `x=5.321`
• `x=5.607`
• `x=5.878`
• `x=6.144`

Graph of `f(x)`:

graph{5x(cosx)^2-10sin(x^2) [0, 6.283, -15, 40]}