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

Answer:

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.

Answer: 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