Graph $\ f(x)=\dfrac{x}{\sqrt{x^{2}+4}}$.

Solution

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 $f$.