Summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of $$ f(x)=\frac{8 x}{x^{2}-1} $$ (Type your answer in interval notation. Use a comma to separate answers as needed.) A. The function is concave upward on $$ \square $$. It is never concave downward. B. The function is concave upward on $$ \square $$. It is concave downward on C. The function is concave downward on $$ \square $$. It is never concave upward. Find the location of any inflection points of $$ f(x) $$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. There is an inflection point at $$ x=\square $$. (Type an integer or a decimal. Use a comma to separate answers as needed.) B. There are no inflection points.

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We begin with the function $$ f(x)=\frac{8x}{x^2-1}. $$ Because the denominator is zero when $$x^2-1=0$$, the domain is $$ (-\infty,-1)\cup(-1,1)\cup(1,\infty). $$ To study concavity we compute the second derivative. First, we compute the first derivative using the quotient rule. Write $$ f(x)=\frac{8x}{x^2-1}. $$ The derivative is $$ f'(x)=\frac{8(x^2-1)-8x(2x)}{(x^2-1)^2}=\frac{8x^2-8-16x^2}{(x^2-1)^2} =\frac{-8x^2-8}{(x^2-1)^2} =-\frac{8(x^2+1)}{(x^2-1)^2}. $$ Now, we differentiate $$f'(x)$$ to find $$f''(x)$$. Write $$ f'(x)=-8\,(x^2+1)(x^2-1)^{-2}. $$ Denote $$ u(x)=x^2+1 \quad \text{and} \quad v(x)=(x^2-1)^{-2}. $$ Then, $$u'(x)=2x$$ and using the chain rule $$ v'(x)=-2\,(x^2-1)^{-3}(2x)=-4x\,(x^2-1)^{-3}. $$ By the product rule we have $$ f''(x)=-8\left[u'(x)v(x)+u(x)v'(x)\right] =-8\left[\frac{2x}{(x^2-1)^2} -\frac{4x(x^2+1)}{(x^2-1)^3}\right]. $$ To combine these terms, express them over the common denominator $$(x^2-1)^3$$: $$ f''(x)=-8\left[\frac{2x(x^2-1)-4x(x^2+1)}{(x^2-1)^3}\right]. $$ Factor $$2x$$ from the numerator: $$ f''(x)=-8\left[\frac{2x\Bigl((x^2-1)-2(x^2+1)\Bigr)}{(x^2-1)^3}\right] =-8\left[\frac{2x\Bigl(x^2-1-2x^2-2\Bigr)}{(x^2-1)^3}\right]. $$ Simplify inside the parentheses: $$ x^2-1-2x^2-2=-x^2-3. $$ Thus, $$ f''(x)=-8\left[\frac{2x(-x^2-3)}{(x^2-1)^3}\right] =\frac{16x(x^2+3)}{(x^2-1)^3}. $$ Since $$x^2+3>0$$ for all $$x$$, the sign of $$f''(x)$$ is determined by the sign of $$x$$ and the cube of $$x^2-1$$. We now analyze the sign on each interval: 1. $$\boldsymbol{x\in(-\infty,-1)}$$: - $$x<0$$ so the factor $$x$$ is negative. - For $$x^2-1$$: if $$x<-1$$, then $$x^2>1$$ making $$x^2-1>0$$; its cube is positive. - Thus, $$f''(x)<0$$ and the graph is concave downward. 2. $$\boldsymbol{x\in(-1,1)}$$: Since the interval must be split at $$x=0$$: - For $$x\in(-1,0)$$: $$x<0$$ (negative) and $$x^2-1<0$$ (since $$x^2<1$$); the cube of a negative number is negative. Hence, $$f''(x)$$ is the quotient of two negatives (a negative $$x$$ and a negative denominator), so $$f''(x)>0$$. The graph is concave upward. - For $$x\in(0,1)$$: Here $$x>0$$ (positive) but $$x^2-1<0$$ (denominator negative), so $$f''(x)<0$$; the graph is concave downward. 3. $$\boldsymbol{x\in(1,\infty)}$$: - $$x>0$$ (positive) and $$x^2-1>0$$ (since $$x^2>1$$); its cube is positive. - Thus, $$f''(x)>0$$; the graph is concave upward. In summary: - The function is concave downward on $$\displaystyle (-\infty,-1)\cup(0,1)$$. - The function is concave upward on $$\displaystyle (-1,0)\cup(1,\infty)$$. Regarding inflection points, look for points in the domain where the concavity changes. In the interval $$(-1,1)$$, the concavity changes at $$x=0$$ (from upward on $$(-1,0)$$ to downward on $$(0,1)$$). Since $$f(0)=\frac{8\cdot 0}{0^2-1}=0$$ and the function is continuous at $$x=0$$, there is an inflection point at $$x=0$$. Thus, the answers are: B. The function is concave upward on $$\displaystyle (-1,0)\cup(1,\infty)$$. It is concave downward on $$\displaystyle (-\infty,-1)\cup(0,1)$$. and A. There is an inflection point at $$x=0$$. The function $$ f(x) = \frac{8x}{x^2 - 1} $$ is concave upward on the intervals $$(-1, 0)$$ and $$(1, \infty)$$, and concave downward on $$(-\infty, -1)$$ and $$(0, 1)$$. There is an inflection point at $$ x = 0 $$.

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