f(x) = x √ (x^2+9) defined on the interval −4≤ x≤5. What is the inflection point? The minimum and maximum occur at x=? f(x) is concave down x= to x=? f(x) is concave up x= to x=?

Answer 1

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

f'(x) = u'v + v'u = (1)(x²+9)^1/2 + (x)(x/((x²+9)^1/2) = (2x² + 9)/((x²+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)/v² = (2x³ + 27x)/(x²+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