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

Question

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

Cameron Alexander · Accepted Answer

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:

Graph of #f(x)#:

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

Paisley Johnson · Answer

undefined To find the points of inflection, first find the second derivative of the function. Then, solve for where the second derivative equals zero and check the concavity of the function around those points.

The second derivative of $ f(x) = 5x\cos^2(x) - 10\sin^2(x) $ is:

$ f''(x) = -20\sin(x)\cos(x) - 20\sin(x)\cos(x) - 10\cos(2x) $

Next, find where the second derivative equals zero:

$ -20\sin(x)\cos(x) - 10\cos(2x) = 0 $

$ -20\sin(x)\cos(x) - 10(2\cos^2(x) - 1) = 0 $

$ -20\sin(x)\cos(x) - 20\cos^2(x) + 10 = 0 $

$ -20\sin(x)\cos(x) - 20(1 - \sin^2(x)) + 10 = 0 $

$ -20\sin(x)\cos(x) - 20 + 20\sin^2(x) + 10 = 0 $

$ -20\sin(x)\cos(x) + 20\sin^2(x) - 10 = 0 $

$ -20\sin(x)(\cos(x) - \sin(x)) - 10 = 0 $

$ -20\sin(x)\sin(x - \pi/4) - 10 = 0 $

There are no solutions for $ x $ in the interval $ [0,2\pi] $. Hence, there are no points of inflection within this interval.