Function concave

Question

Answer the following questions for the function

` f(x) = x \sqrt(x^2 + 36)`

defined on the interval `-6 le x le 6 `.

` f(x) ` is concave down on the interval x = __________________  to x = __________________ 

`f(x)` is concave up on the interval x = __________________  to x = __________________ 

The inflection point for this function is at x = __________________ 

The minimum for this function occurs at x = __________________ 

The maximum for this function occurs at x = __________________ 

Answer 1

The function is odd. It is positive for x > 0 and negative for x < 0, It has 2 horizontal asymptotes of y = -1 (for x negative) and y = 1 (for x positive).

The first derivative is 36/(x^2+36)^3/2, always positive, so the function is monotone increasing. The function is C^∞ and, in particular, C².

Thus, on [-6, 6], the function maximum is at x=6 and its minimum is at -6.

As (x^2+36) is monotone decreasing for x negative and monotone increasing for x positive, the first derivative, 36/(x^2+36)^3/2, is monotone increasing for x negative and monotone decreasing for x positive.

Therefore, we have an inflection point at 0, and the function is concave up for x negative and concave down for x positive.

If you wish, you can calculate f'' = -54x/(x^2+36)^(5/2), which is, as we have determined from looking at the first derivative, positive for x< 0, 0 for x = 0, and negative for x >0, so the function is concave up for negative x, has an inflection at 0, and concave down for x > 0.

Thus, the answers are

concave down on (0, 6]

concave up on [-6, 0)

inflection point at x = 0

min at x = -6

max at x = 6

With a little thought about this function, that it's odd and monotone increasing with horizontal asymptotes, could have led us to this solution.