`f(x) = 24/(x^2 + 12)` Determine the open intervals on whcih the graph is concave upward or downward.

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Chapter 3, 3.4 - Problem 7 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Hello!

To determine where a function is concave upward or downward, we may use the second derivative. This function is defined everywhere, differentiable and second differentiable everywhere, so Concavity Theorem is applicable.

`f'(x) = -24*(2x)/(x^2+12)^2 = -48*(x)/(x^2+12)^2,`

`f''(x) = -48*((x^2+12)^2 - x*[(x^2+12)^2]')/(x^2+12)^4 =`

`-48*((x^2+12)^2 - x*2*(x^2+12)*2x)/(x^2+12)^4 =`

`-48*(x^2+12 - 4x^2)/(x^2+12)^3 =-48*3*(4 - x^2)/(x^2+12)^3 =`

`= 48*3*(x^2-4)/(x^2+12)^3.`

We see that f''(x)=0 for x=-2 and x=2. Moreover, it is negative inside (-2, 2) and is positive outside (the denominator is always positive).

The answer: f is concave upward at `(-oo, -2) uuu(2, +oo)` and is concave downward at (-2, 2).

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