# What are the points of inflection of f(x)=8x^2sin(x-pi/2)  on  x in [0, 2pi]?

Three Points of inflection:

${x}_{1} = 0.59974142102781$
${y}_{1} = - 2.3753382855308$

${x}_{2} = 2.6889674991807$
${y}_{2} = 52.0195659894$

${x}_{3} = 5.385729654028$
${y}_{3} = - 144.70578562542$

#### Explanation:

From the given function
$f \left(x\right) = 8 {x}^{2} \cdot \sin \left(x - \frac{\pi}{2}\right)$
Differentiate to find $f ' \left(x\right)$
$f ' \left(x\right) = 8 \left[2 x \cdot \sin \left(x - \frac{\pi}{2}\right) + {x}^{2} \cdot \cos \left(x - \frac{\pi}{2}\right)\right]$

We are solving for point of inflections, so we take the second derivative $f ' ' \left(x\right)$

$f ' ' \left(x\right) = 16 \left[\sin \left(x - \frac{\pi}{2}\right) + x \cdot \cos \left(x - \frac{\pi}{2}\right)\right] + 8 \left[2 x \cdot \cos \left(x - \frac{\pi}{2}\right) - {x}^{2} \cdot \sin \left(x - \frac{\pi}{2}\right)\right]$

Set $f ' ' \left(x\right) = 0$, then solve for $x$

Simplify first, then solve for $x$

$\sin \left(x - \frac{\pi}{2}\right) \left(16 - 8 {x}^{2}\right) + \cos \left(x - \frac{\pi}{2}\right) \left(16 x + 16 x\right) = 0$
$\tan \left(x - \frac{\pi}{2}\right) = \frac{4 x}{{x}^{2} - 2}$

Using graphics calculator,

${x}_{1} = 0.59974142102781$
and from $f \left(x\right) = 8 {x}^{2} \cdot \sin \left(x - \frac{\pi}{2}\right)$

${y}_{1} = f \left({x}_{1}\right) = 8 {\left(0.59974142102781\right)}^{2} \cdot \sin \left(0.59974142102781 - \frac{\pi}{2}\right)$
${y}_{1} = - 2.3753382855308$

There are other points of inflection $\left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$

${x}_{2} = 2.6889674991807$
${y}_{2} = 52.0195659894$

${x}_{3} = 5.385729654028$
${y}_{3} = - 144.70578562542$

Kindly see the graph of $f \left(x\right) = 8 {x}^{2} \cdot \sin \left(x - \frac{\pi}{2}\right)$ and from $\left[0 , 2 \pi\right]$, there are 3 points of inflection.

graph{y=8x^2*sin(x-pi/2)[-60,60,-30,30]}

God bless...I hope the explanation is useful.