interval question

[Ryan Rigdon]

my problem is

use the Concavity Thm to determine the intervals where the given function is concave up & down. Find all inflection points.

f(x) = 2x^2 + cos^(2)x


my work

f'(x) = 4x - 2cosxsinx

f''(x) = 4 + 2sin^2 (X) - 2cos^2 (x) = 4 + 2(1-cos^2 (x)) - 2cos^2 (x) = 6 - 4cos^2 (x)

6 - 4cos^2 (x) > 0 for all x since 0 (less than or = to) cos^2 (x) (less than or = to) 1

Since f''(x) > 0 concave up for all x. No inflection points.

I am having trouble in writing the interval when its concave up. Would it be [0,1]?

 

[galactus]

It's concave up everywhere.

Then, you could write it is concave up ove rthe interval \(\displaystyle (-\infty, \;\ \infty)\).

Also, the second derivative is >0 for all x, so it is concave up on said interval.

 

[Ryan Rigdon]

So anytime i get f''(x) > 0 concave up for all x. the interval will be (-infinity, infinity). Just want to make sure i understand that. Thanx galactus.

 

[galactus]

Yeah. Unless there is a given interval. Sometimes there is. But this one does not have that.

Take for instance y=x^2. This is conave up over the interval (-inf, inf) as well. Your graph looks a lot like it.

It is concave up if the second derivative is >0 over the interval (a,b). If that interval (a,b) happens to be everywhere, then (-inf,inf).