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

Jul 7, 2017

#### Explanation:

Point of inflection of $f \left(x\right) = x \sin x$ is where an increasing slope starts decreasing or vice-versa. At this point second derivative $\frac{{d}^{2} f \left(x\right)}{{\mathrm{dx}}^{2}} = 0$.

As such using product formula $f \left(x\right) = x \sin x$,

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \sin x + x \cos x$ and

$\frac{{d}^{2} f \left(x\right)}{{\mathrm{dx}}^{2}} = \cos x + \cos x - x \sin x = 2 \cos x - x \sin x$

Now $2 \cos x - x \sin x = 0$ i.e. $x \sin x = 2 \cos x$

or $x = 2 \cot x$

and solution is given by the points where the function $x - 2 \cot x$ cuts $x$-axis.
graph{x-2cotx [-2, 8, -2.5, 2.5]}

Below we give the graph of $f \left(x\right) = x \sin x$ and observe that at these points slope of the curve changes accordingly.
graph{xsinx [-2, 8, -5, 5]}