What are the points of inflection, if any, of f(x)=x+sin^2x  on [0,2pi]?

Mar 17, 2016

They are $\left(\frac{\pi}{4} , \frac{\pi + 2}{4}\right)$, $\left(\frac{3 \pi}{4} , \frac{3 \pi + 2}{4}\right)$, $\left(\frac{5 \pi}{4} , \frac{5 \pi + 2}{4}\right)$, and $\left(\frac{7 \pi}{4} , \frac{7 \pi + 2}{4}\right)$,

Explanation:

Points of inflection are points on the graph at which the concavity changes.

For this function, the concavity changes exactly where the sign of the second derivative changes.

$f \left(x\right) = x + {\sin}^{2} x$

$f ' \left(x\right) = 1 + 2 \sin x \cos x = 1 + \sin \left(2 x\right)$

$f ' ' \left(x\right) = 2 \cos \left(2 x\right)$.

To investigate the sign of the second derivative, find the points at which the sign might change. (In general these are zeros and discontinuities, but this $f ' '$ has no discontinuities.)

$2 \cos \left(2 x\right) = 0$

$2 x = \frac{\pi}{2} + \pi k$ $\text{ }$ ($k$ an integer.)

$x = \frac{\pi}{4} + \frac{\pi}{2} k = \frac{\pi}{4} + \frac{2 \pi}{4} k$ $\text{ }$ ($k$ an integer.)

To find values of $x$ in $\left[0 , 2 \pi\right]$, substitute values of $k$ until we get $x$'s outside the interval. (There are other methods, but this is the one I am choosing here.)

$\left.\begin{matrix}k & \text{|" & -1 & 0 & 1 & 2 & 3 & 4 \\ x & "|} & - \frac{\pi}{4} < 0 & \frac{\pi}{4} & \frac{3 \pi}{4} & \frac{5 \pi}{4} & \frac{7 \pi}{4} & \frac{9 \pi}{4} > 2 \pi\end{matrix}\right.$

Checking the sign of $f ' ' \left(x\right) = 2 \cos \left(2 x\right)$.on the intervals defined by these endpoints we see that the sign does indeed change at each opportunity.

{: (bb "Interval", bb"Sign of "f',bb" Concavity"), ([0,pi/4)," " -" ", " "" Down"), ((pi/4, (3pi)/4), " " +, " " " Up"), (((3pi)/4 ,(5pi)/4), " " -, " "" Down"), (((5pi)/4 ,(7pi)/4), " " +, " "" Up"), (((7pi)/4 ,2pi), " " -, " "" Down") :}

Therefore, the inflection points are the points on the graph with

$x = \frac{\pi}{4}$, $\frac{3 \pi}{4}$, $\frac{5 \pi}{4}$, and $\frac{7 \pi}{4}$

To find the $y$ values use $f \left(x\right) = x + {\sin}^{2} x$