1. Problem statement: Analyze the function $f(x)=\frac{3x^2}{x^2-1}$ including domain, intercepts, asymptotes, critical points, and local extrema. 2. Domain: The denominator is zero when $x^2-1=0$, so $x=\pm 1$ are excluded, hence the domain is $\{x\in\mathbb{R}:x\neq \pm 1\}$. 3. Intercepts: The numerator $3x^2=0$ gives $x=0$, so the only real root and the y-intercept is $(0,0)$. 4. Vertical asymptotes: Because the denominator vanishes at $x=\pm 1$ while the numerator is finite and nonzero there, vertical asymptotes occur at $x=1$ and $x=-1$. $$\lim_{x\to 1^-} f(x) = -\infty$$ $$\lim_{x\to 1^+} f(x) = +\infty$$ $$\lim_{x\to -1^-} f(x) = +\infty$$ $$\lim_{x\to -1^+} f(x) = -\infty$$ 5. Horizontal asymptote: Divide numerator and denominator by $x^2$ to get $$f(x)=\frac{3}{1-1/x^2}$$ so $$\lim_{x\to \pm\infty} f(x)=3$$ 6. Derivative and critical points: Use the quotient rule to compute $$f'(x)=\frac{6x(x^2-1)-3x^2(2x)}{(x^2-1)^2}$$ which simplifies to $$f'(x)=-\frac{6x}{(x^2-1)^2}$$ Setting $f'(x)=0$ gives $x=0$ as the only critical point in the domain. 7. Second derivative and classification: Differentiate to obtain $$f''(x)=\frac{6(3x^2+1)}{(x^2-1)^3}$$ Evaluating at $x=0$ yields $f''(0)=\frac{6(1)}{(-1)^3}=-6<0$, so the point $(0,0)$ is a local maximum. 8. Monotonicity and sign: Since $f'(x)=-6x/(x^2-1)^2$, $f'(x)>0$ when $x<0$ and $f'(x)<0$ when $x>0$, so $f$ is increasing on $(-\infty,-1)$ and $(-1,0)$ and decreasing on $(0,1)$ and $(1,\infty)$. For $|x|<1$ the denominator is negative and the numerator nonnegative, so $f(x)\le 0$ on $(-1,1)$ with equality only at $0$, and for $|x|>1$ we have $f(x)>0$. 9. Final summary: Domain $x\neq \pm 1$; vertical asymptotes at $x=\pm 1$; horizontal asymptote $y=3$; intercept and local maximum at $(0,0)$; increasing for $x<0$ and decreasing for $x>0$, with monotonicity split by the asymptotes.