Question

which function has the horizontal asymptotes y = 1 and y = -1? f(x)=\frac{2x^{2}-1}{2x^{2}+x}, f(x)=\frac{1 - x^{2}}{1 + x^{2}}, f(x)=e^{-x}+1, f(x)=\frac{2}{pi}\tan^{-1}x

Explanation:

Step1: Recall horizontal - asymptote formula

The horizontal asymptotes of a function \(y = f(x)\) are found by calculating \(\lim_{x
ightarrow\infty}f(x)\) and \(\lim_{x
ightarrow-\infty}f(x)\).

Step2: Analyze \(f(x)=\frac{1 - x^{2}}{1 + x^{2}}\)

For \(\lim_{x
ightarrow\pm\infty}\frac{1 - x^{2}}{1 + x^{2}}\), divide numerator and denominator by \(x^{2}\): \(\lim_{x
ightarrow\pm\infty}\frac{\frac{1}{x^{2}}-1}{\frac{1}{x^{2}} + 1}\). As \(x
ightarrow\pm\infty\), \(\frac{1}{x^{2}}
ightarrow0\). So \(\lim_{x
ightarrow\pm\infty}\frac{\frac{1}{x^{2}}-1}{\frac{1}{x^{2}} + 1}=- 1\).

Step3: Analyze \(f(x)=\frac{2x^{2}-1}{2x^{2}+x}\)

Divide numerator and denominator by \(x^{2}\): \(\lim_{x
ightarrow\pm\infty}\frac{2-\frac{1}{x^{2}}}{2+\frac{1}{x}}\). As \(x
ightarrow\pm\infty\), \(\frac{1}{x^{2}}
ightarrow0\) and \(\frac{1}{x}
ightarrow0\), so \(\lim_{x
ightarrow\pm\infty}\frac{2-\frac{1}{x^{2}}}{2+\frac{1}{x}} = 1\). But for \(x
ightarrow-\infty\), the behavior is not consistent with having \(y=-1\) as an asymptote.

Step4: Analyze \(f(x)=e^{-x}+1\)

\(\lim_{x
ightarrow\infty}e^{-x}+1 = 1\) and \(\lim_{x
ightarrow-\infty}e^{-x}+1=\infty\), so it doesn't have \(y = - 1\) as an asymptote.

Step5: Analyze \(f(x)=\frac{2}{\pi}\tan^{- 1}x\)

We know that \(\lim_{x
ightarrow\infty}\tan^{-1}x=\frac{\pi}{2}\) and \(\lim_{x
ightarrow-\infty}\tan^{-1}x=-\frac{\pi}{2}\). Then \(\lim_{x
ightarrow\infty}\frac{2}{\pi}\tan^{-1}x = 1\) and \(\lim_{x
ightarrow-\infty}\frac{2}{\pi}\tan^{-1}x=-1\).

Answer:

\(f(x)=\frac{2}{\pi}\tan^{-1}x\)