Question

How do you find the maximum, minimum and inflection points for each function `(x)/(x^2-9)`?

Answer 1

No relative maximum or minimum
Inflection at `(0, 0)`

Explanation:

Given: `f(x) = x/(x^2-9) = x/((x-3)(x+3))`

Find critical values: `f'(x) = 0`:

Find the first derivative using the quotient rule:

`(u/v)' = (v u' - u v')/v^2`

Let `u = x; " "u' = 1`

Let `v = x^2-9; " "v' = 2x`

`f'(x) = ((x^2-9)(1) - x(2x))/(x^2-9)^2 = (x^2 - 2x^2 -9)/(x^2-9)^2 = (-x^2 - 9)/(x^2-9)^2`

`f'(x) = (-(x^2 +9))/(x^2-9)^2 = 0`

`-(x^2 +9) = 0; " "x^2 + 9 = 0; " "=> x = +-3i`

No critical values.

Inflection points are found when `f''(x) = 0`

Find the second derivative:

Let `u = -x^2 - 9; " "u' = -2x`

Let `v = (x^2 - 9)^2; " "v' = 2(x^2-9)(2x) = 4x(x^2-9)`

Find `f''(x) = (((x^2 - 9)^2)(-2x) - (-x^2 - 9)(4x(x^2-9)))/((x^2-9)^4)`

`f''(x) = (((x^2 - 9)^2)(-2x) + (x^2 + 9)(4x(x^2-9)))/((x^2-9)^4)`

`f''(x) = ((x^2-9)[-2x(x^2-9) + 4x(x^2+9)])/((x^2-9)^4)`

`f''(x) = (-2x^3+18x + 4x^3 + 36x)/(x^2-9)^3 = (2x^3 + 54x)/(x^2-9)^3`

`f''(x) = (2x(x^2 + 27))/(x^2-9)^3 = 0`

`2x(x^2 + 27) = 0`

`x = 0, x = +- sqrt(27)i = +- 3sqrt(3)i`

Inflection point at `x = 0`

`f(0) = 0`

Inflection at `(0, 0)`