What are the points of inflection of #f(x)=xcos^2x + x^2sinx
#?

Question

What are the points of inflection of #f(x)=xcos^2x + x^2sinx
  #?

Emily Ammerman · Accepted Answer

The point #(0,0)#.

In order to find the inflection points of #f#, you have to study the variations of #f'#, and to do that you need to derivate #f# two times. #f'(x) = cos^2(x) + x(-sin(2x) + 2sin(x) + xcos(x))# #f''(x) = -2sin(2x) + 2sin(x) + x(-2cos(2x) + 4cos(x) - xsin(x))# The inflection points of #f# are the points when #f''# is zero and goes from positive to negative. #x = 0# seems to be such a point because #f''(pi/2) > 0# and #f''(-pi/2) < 0#

Axel Alvarez · Answer

undefined To find the points of inflection of $ f(x) = x \cos^2(x) + x^2 \sin(x) $, we need to find its second derivative and then determine where it equals zero.

First, let's find the first derivative of $ f(x) $:
$$ f'(x) = \frac{d}{dx}[x \cos^2(x)] + \frac{d}{dx}[x^2 \sin(x)] $$

Using the product rule:
$$ f'(x) = \cos^2(x) - 2x \sin(x) \cos(x) + 2x \sin(x) + 2x \cos^2(x) $$

Now, let's find the second derivative of $ f(x) $:
$$ f''(x) = \frac{d}{dx}[\cos^2(x)] - 2\sin(x)\cos(x) - 2x\cos(x) + 2\sin(x) + 2\cos^2(x) + 2\sin(x) - 2x\sin(x) $$

$$ f''(x) = -2\sin(x)\cos(x) - 2\cos(x) - 2x\sin(x) + 2\sin(x) + 2\cos^2(x) - 2x\sin(x) $$

$$ f''(x) = -2\cos(x)[\sin(x) + 1] - 2x\sin(x) + 2\sin(x) + 2\cos^2(x) $$

To find the points of inflection, we set $ f''(x) = 0 $ and solve for $ x $. However, finding exact solutions for this equation might be challenging, and numerical methods or graphical analysis may be required.

What are the points of inflection of #f(x)=xcos^2x + x^2sinx #?