# Graph f(x)=\frac{x}{(2x-3)^2}. Label in the graph intervals where the function is concave up or...

## Question:

Graph {eq}f(x)=\frac{x}{(2x-3)^2} {/eq}.

Label in the graph intervals where the function is concave up or concave down.

## Concavity:

The concavity of a function is directly related to the second derivative of the function. A function is concave up when its second derivative is positive, and a function is concave down when its second derivative is negative.

## Answer and Explanation:

**To determine the function's concavity, we find and analyze the second derivative:**

{eq}\displaystyle f(x)=\frac{x}{(2x-3)^2} \displaystyle f'(x) = \frac{1\cdot \left(2x-3\right)^2-4\left(2x-3\right)x}{\left(\left(2x-3\right)^2\right)^2}\\ \displaystyle f'(x) = \frac{-3-2x}{\left(2x-3\right)^3}\\ \displaystyle f''(x) = \frac{\left(-2\right)\left(2x-3\right)^3-6\left(2x-3\right)^2\left(-3-2x\right)}{\left(\left(2x-3\right)^3\right)^2}\\ \displaystyle f''(x) = \frac{8\left(x+3\right)}{\left(2x-3\right)^4}\\ \displaystyle f''(x)=0 \,\, \rightarrow \,\, x+3=0\\ \displaystyle \boxed{x=-3} \,\, \rightarrow \,\, \textrm {point where the second derivative is zero} {/eq}

To define the sign of the second derivative in each interval, evaluate a point of each interval and verify the sign:

{eq}\left (-\infty,-3\right) \,\,\,\, \rightarrow \,\,\,\, f''< 0 \,\,\,\,\, \textrm { the second derivative is negative in this interval } \\ \left (-3, \frac{3}{2}\right) \,\,\,\, \rightarrow \,\,\,\, f'' > 0 \,\,\,\,\, \textrm { the second derivative is positive in this interval }\\ \left (\frac{3}{2},\infty \right) \,\,\,\, \rightarrow \,\,\,\, f'' > 0 \,\,\,\,\, \textrm { the second derivative is positive in this interval} {/eq}

**Conclusion (Concavity) ** {eq}\boxed{ \left (-\infty,-3\right)} \,\, \Longrightarrow \,\, \textrm {concave down}\\ \boxed{ \left (-3, \frac{3}{2}\right) \bigcup \left (\frac{3}{2},\infty \right) } \,\, \Longrightarrow \,\, \textrm {concave up}\\ {/eq}

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