# Find the open intervals on which the function f(x) = \frac{1}{3x^2+4} is concave up or down....

## Question:

Find the open intervals on which the function {eq}f(x) = \frac{1}{3x^2+4} {/eq} is concave up or down. Then, determine the x-coordinates of all inflection points of the function.

## Concavity and Inflection Points:

Given a continuous function {eq}F(x){/eq}. The concavity of the function is given by its second derivative such that at any point {eq}x=c{/eq} on its domain, {eq}F(x){/eq}:

• has an inflection point when {eq}F''(c) = 0 {/eq};
• is concave up when {eq}F''(c) > 0{/eq}; and,
• is concave down when {eq}F''(c) < 0{/eq}.

Given the function {eq}f(x) = \frac{1}{3x^2 + 4}{/eq} where the function's domain is the set of all real numbers {eq}R{/eq}.

Determine the inflection points by solving for the zeros of the second derivative of the function:

{eq}\begin{align*} A''(x) &= \frac{d^2}{dx^2} \left( \frac{1}{3x^2 + 4} \right) & \text{[Differentiate with respect to } x \text{]} \\ &= \frac{d}{dx} \left( -\frac{6x}{(3x^2 + 4)^2} \right) & \text{[Differentiate further with respect to } x \text{]} \\ &= \frac{ 54x^2 - 24 }{(3x^2+4)^3} \\ A''(x) &= \frac{ 6 (3x + 2) (3x - 2) }{(3x^2+4)^3} & \text{[Solve for the zeros } of A''(x) \text{]} \\ 0 &= \frac{ 6 (3x + 2) (3x - 2) }{(3x^2+4)^3} \\ 0 &= 6 (3x + 2) (3x - 2) \\ 3x + 2 &= 0 \text{ or } 3x-2 = 0 \\ x = &-\frac{2}{3} \text{ or } x =\frac{2}{3} & \boxed{f(x) \text{ has inflection points at } x \in \left\{ -\frac{2}{3}, \frac{2}{3} \right\} } \end{align*} {/eq}

These inflection points divide the domain of the function into the intervals {eq}\left( -\infty, -\frac{2}{3} \right),\, \left( -\frac{2}{3}, \frac{2}{3} \right) \text{ and } \left( \frac{2}{3}, \infty \right) {/eq}.

Evaluate the sign of {eq}A''(x){/eq} at an arbitrary value from each of the intervals to determine the concavity of the function:

{eq}\begin{align*} &\left( -\infty, -\frac{2}{3} \right) \text{: Let } x = -1 \\ &\Rightarrow A''(-1) = \frac{ 54(-1)^2 - 24 }{(3(-1)^2+4)^3} \\ &\Rightarrow A''(-1) = \frac{ 54 - 24 }{(3+4)^3} \\ &\Rightarrow A''(-1) = \frac{ 30 }{7^3} \\ &\Rightarrow A''(-1) = \frac{ 30 }{343} > 0 & \boxed{f(x) \text{ is concave up on the interval } \left( -\infty, -\frac{2}{3} \right) } \\ \\ &\left( -\frac{2}{3}, \frac{2}{3} \right) \text{: Let } x = 0 \\ &\Rightarrow A''(0) = \frac{ 54(0)^2 - 24 }{(3(0)^2+4)^3} \\ &\Rightarrow A''(0) = \frac{ - 24 }{4^3} \\ &\Rightarrow A''(0) = -\frac{3}{8} < 0 & \boxed{f(x) \text{ is concave down on the interval } \left( -\frac{2}{3}, \frac{2}{3} \right) } \\ \\ &\left( \frac{2}{3}, \infty \right) \text{: Let } x = 1 \\ &\Rightarrow A''(1) = \frac{ 54(1)^2 - 24 }{(3(1)^2+4)^3} \\ &\Rightarrow A''(1) = \frac{ 54 - 24 }{(3+4)^3} \\ &\Rightarrow A''(1) = \frac{ 30 }{7^3} \\ &\Rightarrow A''(1) = \frac{ 30 }{343} > 0 & \boxed{f(x) \text{ is concave up on the interval } \left( \frac{2}{3}, \infty \right) } \end{align*} {/eq}