Question

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

Answer 1

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(x)=8x^2*sin(x-pi/2)`
Differentiate to find `f' (x)`
`f' (x)=8[2x*sin(x-pi/2)+x^2*cos(x-pi/2)]`

We are solving for point of inflections, so we take the second derivative `f''(x)`

`f'' (x)=16[sin(x-pi/2)+x*cos(x-pi/2)]+8[2x*cos(x-pi/2)-x^2*sin(x-pi/2)]`

Set `f'' (x)=0`, then solve for `x`

Simplify first, then solve for `x`

`sin(x-pi/2)(16-8x^2)+cos(x-pi/2)(16x+16x)=0`
`tan(x-pi/2)=(4x)/(x^2-2)`

Using graphics calculator,

`x_1=0.59974142102781`
and from `f(x)=8x^2*sin(x-pi/2)`

`y_1=f(x_1)=8(0.59974142102781)^2*sin(0.59974142102781-pi/2)`
`y_1=-2.3753382855308`

There are other points of inflection `(x_2, y_2)` and `(x_3, y_3)`

`x_2=2.6889674991807`
`y_2=52.0195659894`

`x_3=5.385729654028`
`y_3=-144.70578562542`

Kindly see the graph of `f(x)=8x^2*sin(x-pi/2)` and from `[0, 2pi]`, 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.