The Q (AND MY Q, IN REGARDS TO THE ANSWER WILL BE FOUND BELOW):

If n is a positive integer, how many times does the function f(x)=x^2 + 5cosx change concavity in the interval?

0 less than or equal to x less than or equal to 2(pi)*n

f '(x) = 2x - 5*sin(x)

Set this equal to 0

x = ((5*sin(x)) / 2)

Take the second derivative.

f"(x) = -5*cos(x)

This will be a point of inflection for x = π/2, 3π/2, 5π/2...

So there will be a point of inflection every π increase. For 2π that will be 2, and for 2π*n there will be 2n.

MY Q: How is the second derivitive -5*cos(x). Why isn't it plus 2? And is the X solved for so that it can be plugged in to the inequality to detirmine how long it takes for "n" to have an inflection point? (like it cycles every 2pn basically)

thanks

Concavity Theorem:
If the function $f(x)$ is twice differentiable at $x = c$ , then the graph of $f(x)$ is concave upward at $(c,f(c))$
if $f''(c)>0$ and concave downward if $f''(c)<0$

Now we know that:
$f(x)=x^2 + 5\cos(x)$

To find the concavity we take the second derivative of the function, equal it to zero and find the values of $x$. Those values of $x$ are the points(inflection points) where concavity changes.

So:
\begin{align*} f''(x) = 2 - 5\cos(x) =& 0\\ 5\cos(x) =& 2....\text{[using simple Algebra]}\\ x =& \cos^{-1}\left(\frac{2}{5}\right)\\ x =& 66.42^{\circ}, 360^\circ - 66.42^\circ , 360^\circ +66.42^\circ , (360^\circ * 2) - 66.42^\circ, ......\end{align*}

So the function changes concavity $2n$ times where $n = 0, 1, 2, 3, .....$ within the interval $[0, n*2\pi]$.

You can see the plot of the function here:

plot x^2 + 5cos(x), x = 0 to 720 degree - Wolfram|Alpha