# How do you determine the number of inflection points

• May 4th 2009, 09:06 PM
swatpup32
How do you determine the number of inflection points
for the function $y=2x+cos(x^2)$ on the interval of $0\leq x \leq 5$
$f\prime\prime = -2x(2xcosx^2)- 2sinx^2$

I don't need to find the inflection values. Just the number of inflections points for the graph over the interval.
• May 4th 2009, 10:09 PM
redsoxfan325
Quote:

Originally Posted by swatpup32
for the function $y=2x+cos(x^2)$ on the interval of $0\leq x \leq 5$
$f\prime\prime = -2x(2xcosx^2)- 2sinx^2$

I don't need to find the inflection values. Just the number of inflections points for the graph over the interval.

Solve $-4x^2\cos(x^2)-2\sin(x^2)=0$ on $[0,5]$

Let $u=x^2$ (this extends the domain to $[0,5^2]$)

$-4u\cos(u)-2\sin(u)=0 \implies 2u\cos(u)+\sin(u)=0$.

Clearly, a solution is $u=0 \implies x=0$. There are eight others, but I only know that because I graphed the equation and counted them. I honestly don't know how to solve for the other roots analytically.

But there are 9 zeroes and thus 9 inflection points.
• May 5th 2009, 06:27 AM
Kaitosan
Quote:

Originally Posted by redsoxfan325
But there are 9 zeroes and thus 9 inflection points.

Not neccessarily true. An inflection point is a point in which (1.) the slope of f' is zero or undefined vertically and (2.) there is a change of the sign of the slope of f' from - to + or from + to - . A zero can exist without the slope sign changing, therefore it'd not be an inflection point.
• May 5th 2009, 07:42 AM
redsoxfan325
Ok, but this is a continuous function defined on a compact set, so it can't be unbounded, and whenever it changes from positive to negative there will be a corresponding zero by the IVT.
• May 5th 2009, 02:09 PM
TheEmptySet
I have an indeterminate form ie $0\cdot \infty$ this is not well defined, and I checked if the limits go to zero and they do not.

The zero factor principle does not apply becuase $0\cdot \infty \ne =$

Thanks for the graph skeeter.

So the 2nd derivative is

$f''(x)=-4x^2\cos(x^2)-2\sin(x^2)$

This can be factored as follows

$0=-2\cos(x^2)[2x^2+\tan(x^2)]$

The 2nd factor only has x=0 as a solution so the solutions are

$\cos(x^2)=0 \iff x^2=\frac{\pi}{2}+n \pi= \frac{(2n+1) \pi}{2}$

$x=\pm \sqrt{\frac{(2n+1)\pi}{2}}$

Note that in the factorization that $(2x^2+\tan(x^2)) > 0$ for all $x> 0$

Since $\cos(x^2)$ changes sign at each of it zero all of the above zero's are critical points
• May 5th 2009, 02:31 PM
skeeter
graph of the 2nd derivative ...