Find the domain and the vertical and horizontal asymptotes. Separate your asymptote answers with commas, if there are multiple, or enter "DNE" if there are none. \( f(x)=\frac{x+1}{x^{2}+5} \) Domain: Vertical asymptote(s): \( x= \) Horizontal asymptote(s): \( y= \)

Analyze the critical points of the function \( f(x)=\frac{x+1}{x^{2}+5} \) Function by following steps: - step0: Find the critical numbers: \(f\left(x\right)=\frac{x+1}{x^{2}+5}\) - step1: Find the domain: \(f\left(x\right)=\frac{x+1}{x^{2}+5},x \in \mathbb{R}\) - step2: Find the derivative: \(f^{\prime}\left(x\right)=\frac{-x^{2}+5-2x}{\left(x^{2}+5\right)^{2}}\) - step3: Find the domain: \(f^{\prime}\left(x\right)=\frac{-x^{2}+5-2x}{\left(x^{2}+5\right)^{2}},x \in \mathbb{R}\) - step4: Substitute \(f^{\prime}\left(x\right)=0:\) \(0=\frac{-x^{2}+5-2x}{\left(x^{2}+5\right)^{2}}\) - step5: Swap the sides: \(\frac{-x^{2}+5-2x}{\left(x^{2}+5\right)^{2}}=0\) - step6: Cross multiply: \(-x^{2}+5-2x=\left(x^{2}+5\right)^{2}\times 0\) - step7: Simplify the equation: \(-x^{2}+5-2x=0\) - step8: Rewrite in standard form: \(-x^{2}-2x+5=0\) - step9: Multiply both sides: \(x^{2}+2x-5=0\) - step10: Solve using the quadratic formula: \(x=\frac{-2\pm \sqrt{2^{2}-4\left(-5\right)}}{2}\) - step11: Simplify the expression: \(x=\frac{-2\pm \sqrt{24}}{2}\) - step12: Simplify the expression: \(x=\frac{-2\pm 2\sqrt{6}}{2}\) - step13: Separate into possible cases: \(\begin{align}&x=\frac{-2+2\sqrt{6}}{2}\\&x=\frac{-2-2\sqrt{6}}{2}\end{align}\) - step14: Simplify the expression: \(\begin{align}&x=-1+\sqrt{6}\\&x=\frac{-2-2\sqrt{6}}{2}\end{align}\) - step15: Simplify the expression: \(\begin{align}&x=-1+\sqrt{6}\\&x=-1-\sqrt{6}\end{align}\) Analyze the extrema of the function \( f(x)=\frac{x+1}{x^{2}+5} \) Function by following steps: - step0: Find the local extrema: \(f\left(x\right)=\frac{x+1}{x^{2}+5}\) - step1: Find the domain: \(f\left(x\right)=\frac{x+1}{x^{2}+5},x \in \mathbb{R}\) - step2: Find the derivative: \(f^{\prime}\left(x\right)=\frac{-x^{2}+5-2x}{\left(x^{2}+5\right)^{2}}\) - step3: Find the domain: \(f^{\prime}\left(x\right)=\frac{-x^{2}+5-2x}{\left(x^{2}+5\right)^{2}},x \in \mathbb{R}\) - step4: Substitute \(f^{\prime}\left(x\right)=0:\) \(0=\frac{-x^{2}+5-2x}{\left(x^{2}+5\right)^{2}}\) - step5: Swap the sides: \(\frac{-x^{2}+5-2x}{\left(x^{2}+5\right)^{2}}=0\) - step6: Cross multiply: \(-x^{2}+5-2x=\left(x^{2}+5\right)^{2}\times 0\) - step7: Simplify the equation: \(-x^{2}+5-2x=0\) - step8: Rewrite in standard form: \(-x^{2}-2x+5=0\) - step9: Multiply both sides: \(x^{2}+2x-5=0\) - step10: Solve using the quadratic formula: \(x=\frac{-2\pm \sqrt{2^{2}-4\left(-5\right)}}{2}\) - step11: Simplify the expression: \(x=\frac{-2\pm \sqrt{24}}{2}\) - step12: Simplify the expression: \(x=\frac{-2\pm 2\sqrt{6}}{2}\) - step13: Separate into possible cases: \(\begin{align}&x=\frac{-2+2\sqrt{6}}{2}\\&x=\frac{-2-2\sqrt{6}}{2}\end{align}\) - step14: Simplify the expression: \(\begin{align}&x=-1+\sqrt{6}\\&x=\frac{-2-2\sqrt{6}}{2}\end{align}\) - step15: Simplify the expression: \(\begin{align}&x=-1+\sqrt{6}\\&x=-1-\sqrt{6}\end{align}\) - step16: Determine the intervals: \(\begin{align}&\left(-1-\sqrt{6},-1+\sqrt{6}\right),\left(-1+\sqrt{6},+\infty\right)\\&\left(-\infty,-1-\sqrt{6}\right),\left(-1-\sqrt{6},-1+\sqrt{6}\right)\end{align}\) - step17: Choose the points: \(\begin{align}&x_{1}=-1,x_{2}=2\\&x_{3}=-4,x_{4}=-1\end{align}\) - step18: Find the values of the derivatives: \(\begin{align}&f^{\prime}\left(-1\right)=\frac{1}{6},f^{\prime}\left(2\right)=-\frac{1}{27}\\&f^{\prime}\left(-4\right)\approx -0.006803,f^{\prime}\left(-1\right)=\frac{1}{6}\end{align}\) - step19: The local maximum is at \(x=-1+\sqrt{6}:\) \(\begin{align}&f\left(x\right)=\frac{x+1}{x^{2}+5},x=-1+\sqrt{6}\\&f\left(x\right)=\frac{x+1}{x^{2}+5},x=-1-\sqrt{6}\end{align}\) - step20: Calculate the function value: \(\begin{align}&f\left(-1+\sqrt{6} \right)=\frac{\sqrt{6}+1}{10}\\&f\left(-1-\sqrt{6} \right)=\frac{1-\sqrt{6}}{10}\end{align}\) - step21: The local maximum of the function is \(\frac{\sqrt{6}+1}{10}\) at \(x = -1+\sqrt{6}:\) \(\begin{align}&\textrm{The local maximum is }\frac{\sqrt{6}+1}{10}\textrm{ at }x = -1+\sqrt{6}\\&\textrm{The local minimum is }\frac{1-\sqrt{6}}{10}\textrm{ at }x = -1-\sqrt{6}\end{align}\) The domain of the function \( f(x)=\frac{x+1}{x^{2}+5} \) is all real numbers except for the values where the denominator is equal to zero. The critical points are \( x=-1+\sqrt{6} \) and \( x=-1-\sqrt{6} \). The local maximum is \( \frac{\sqrt{6}+1}{10} \) at \( x=-1+\sqrt{6} \) and the local minimum is \( \frac{1-\sqrt{6}}{10} \) at \( x=-1-\sqrt{6} \). To find the vertical asymptotes, we need to determine the values of \( x \) for which the denominator \( x^{2}+5 \) is equal to zero. The vertical asymptotes occur at these values. To find the horizontal asymptote, we need to analyze the behavior of the function as \( x \) approaches positive or negative infinity. The horizontal asymptote is the value that the function approaches as \( x \) goes to positive or negative infinity. Let's calculate the vertical asymptotes and the horizontal asymptote. Function by following steps: - step0: Find the vertical asymptotes: \(f\left(x\right)=\frac{x+1}{x^{2}+5}\) - step1: The function has no vertical asymptote: \(\textrm{No vertical asymptotes}\) Find the horizontal asymptotes of the function \( f(x)=\frac{x+1}{x^{2}+5} \) Function by following steps: - step0: Find the horizontal asymptotes: \(f\left(x\right)=\frac{x+1}{x^{2}+5}\) - step1: Evaluate the limits \(\lim _{x\rightarrow +\infty}\left(f\left(x\right)\right)\) and \(\lim _{x\rightarrow -\infty}\left(f\left(x\right)\right):\) \(\begin{align}&\lim _{x\rightarrow +\infty}\left(\frac{x+1}{x^{2}+5}\right)\\&\lim _{x\rightarrow -\infty}\left(\frac{x+1}{x^{2}+5}\right)\end{align}\) - step2: Calculate: \(\begin{align}&0\\&0\end{align}\) - step3: The finite values are horizontal asymptotes: \(\begin{align}&f\left(x\right)=0\end{align}\) The domain of the function \( f(x)=\frac{x+1}{x^{2}+5} \) is all real numbers. The vertical asymptotes are not present for this function. The horizontal asymptote is \( y=0 \). Therefore, the domain is all real numbers, the vertical asymptotes are not present, and the horizontal asymptote is \( y=0 \).