Graph \(\ f(x)=\dfrac{x}{\sqrt{x^{2}+4}}\).
Step 1 The domain of \(f\) is all real numbers. The only intercept is \((0,0)\). So, we plot the point \((0,0) .\)
Step 2 We check for horizontal asymptotes. Since \[ \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }f(x) =\lim\limits_{x\rightarrow \infty }\dfrac{x}{\sqrt{x^{2}+4}}=\sqrt{ \lim\limits_{x\rightarrow \infty }\dfrac{x^{2}}{x^{2}+4}} \underset{\underset{{\color{#0066A7}{\hbox{L’Hôpital’s Rule}}}} {\color{#0066A7}{{{\left\uparrow{\vphantom{width0pc height9pt depth0pt}}\right.}}}}}{=} \sqrt{\lim\limits_{x\rightarrow \infty }\dfrac{2x}{2x}}=1\\[-12.9pt] \end{eqnarray*} \]
the line \(y=1\) is a horizontal asymptote as \(x\rightarrow \infty .\) Since \(\lim\limits_{x\rightarrow -\infty }f(x) =-1,\) the line \(y=-1\) is also a horizontal asymptote as \(x\rightarrow -\infty .\) We draw the horizontal asymptotes on the graph.
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Since \(f\) is defined for all real numbers, there are no vertical asymptotes.
Step 3\[ \begin{array}{@{\hspace*{-2.6pc}}rcl} \!\!f^\prime (x) &=&\dfrac{d}{\textit{dx}}\left( \dfrac{x }{\sqrt{x^{2}+4}}\right) =\dfrac{1\cdot \sqrt{x^{2}+4}-x\left[ \dfrac{1}{2} ( x^{2}+4) ^{-1/2}\cdot 2x\right] }{x^{2}+4}\\[5pt] &=&\dfrac{\sqrt{x^{2}+4}-\dfrac{x^{2}}{\sqrt{x^{2}+4}}}{x^{2}+4} =\dfrac{x^{2}+4-x^{2}}{( x^{2}+4) \sqrt{x^{2}+4}}=\dfrac{4}{(x^{2}+4)^{3/2}}\\ f^{\prime \prime} (x) &=& \dfrac{d}{\textit{dx}} \left[ \dfrac{4}{( x^{2}+4) ^{3/2}}\right] =4\left[ -\dfrac{3}{2} ( x^{2}+4) ^{-5/2}\right] \cdot 2x=\dfrac{-12x}{( x^{2}+4) ^{5/2}} \end{array} \]
Since \(f^\prime (x)\) is never \(0\) and \(f^\prime\) exists for all real numbers, there are no critical numbers.
Step 4 Since \(f^\prime (x) >0\) for all \(x\), \(f\) is increasing on \(( -\infty ,\infty)\) .
Step 5 Because there are no critical numbers, there are no local extreme values.
Step 6 We test for concavity. Since \(f^{\prime \prime} (x) >0\) on the interval \((-\infty ,0) \), \(f\) is concave up on \((-\infty ,0)\); and since \(f^{\prime \prime} (x) <0\) on the interval \(( 0,\infty ) \), \(f\) is concave down on \(( 0,\infty ) \). The concavity changes at \(0,\) and \(0\) is in the domain of \(f\), so the point \((0,0)\) is an inflection point of \(f\). Plot the inflection point.
Step 7 Figure 49 shows the graph of \(f\).