What are the points of inflection of #f(x)=xsinx # on the interval #x in [0,2pi]#?

Question

Emilia Aguilar · Accepted Answer

Please see below.

Point of inflection of #f(x)=xsinx# is where an increasing slope starts decreasing or vice-versa. At this point second derivative #(d^2f(x))/(dx^2)=0#. As such using product formula #f(x)=xsinx#, #(df(x))/(dx)=sinx+xcosx# and #(d^2f(x))/(dx^2)=cosx+cosx-xsinx=2cosx-xsinx# Now #2cosx-xsinx=0# i.e. #xsinx=2cosx# or #x=2cotx# and solution is given by the points where the function #x-2cotx# cuts #x#-axis. graph{x-2cotx [-2, 8, -2.5, 2.5]} Below we give the graph of #f(x)=xsinx# and observe that at these points slope of the curve changes accordingly. graph{xsinx [-2, 8, -5, 5]}

Christopher Agresta · Answer

undefined To find the points of inflection of $ f(x) = x \sin(x) $ on the interval $ x $ in $[0,2\pi]$, we first need to find the second derivative of $ f(x) $, then find the values of $ x $ where the second derivative changes sign from positive to negative or vice versa.

First derivative: $ f'(x) = x \cos(x) + \sin(x) $

Second derivative: $ f''(x) = -x \sin(x) + 2 \cos(x) $

Now, we need to find where $ f''(x) = 0 $ or does not exist, and then determine the sign of $ f''(x) $ around those points.

Setting $ f''(x) = 0 $, we solve $ -x \sin(x) + 2 \cos(x) = 0 $. This equation doesn't have solutions in the interval $ [0,2\pi] $.

Next, we determine the sign of $ f''(x) $ in the intervals determined by the critical points of $ f''(x) $.

We observe the sign changes of $ f''(x) $ around the critical points. These sign changes indicate points of inflection.

Analyzing the sign of $ f''(x) $ around the critical points in the interval $ [0,2\pi] $, we find that there are points of inflection at $ x = \frac{\pi}{2} $ and $ x = \frac{3\pi}{2} $.