# Thread: How do you determine the number of inflection points

1. ## How do you determine the number of inflection points

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

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

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

I don't need to find the inflection values. Just the number of inflections points for the graph over the interval.
Solve $\displaystyle -4x^2\cos(x^2)-2\sin(x^2)=0$ on $\displaystyle [0,5]$

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

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

Clearly, a solution is $\displaystyle u=0 \implies x=0$. 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.

3. 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.

4. 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.

5. I have an indeterminate form ie $\displaystyle 0\cdot \infty$ 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 $\displaystyle 0\cdot \infty \ne =$

Thanks for the graph skeeter.

So the 2nd derivative is

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

This can be factored as follows

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

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

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

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

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

Since $\displaystyle \cos(x^2)$ changes sign at each of it zero all of the above zero's are critical points

6. graph of the 2nd derivative ...