William W. answered • 04/08/19

Experienced Tutor and Retired Engineer

To find the critical points, take the derivative. Using the product rule, the derivative is:

f'(x) = u'v + v'u = (1)(x^{2}+9)^{1/2} + (x)(x/((x^{2}+9)^{1/2}) = (2x^{2} + 9)/((x^{2}+9)^{1/2}) This has no solutions so there are no critical points (except the endpoints at x = -4 and x = 5). The derivative is always positive so the function is always increasing meaning x = -4 is the minimum and x = 5 is the maximum.

To find where the function is concave up and down, take the second derivative. f''(x" = (f'(x))'. Use the quotient rule to find this.

f''(x) = (u'v - v'u)/v^{2} = (2x^{3} + 27x)/(x^{2}+9)^{3/2} This equals zero when x = 0 so x = 0 is an inflection point. -4<x<0 is concave down because f''(x) < 0 on this interval and 0<x<5 is concave up because f''(x) > 0 on this interval