4.5: Graphing Using Calculus - Putting it Altogether

As \(x \to \infty\), \(e^{-2x} \to 0\) leaving us with \(3x - 1\). Thus, our claim is that \(y = 3x - 1\) is a slant asymptote for the function \(f\) as \(x \to \infty\). To prove this statement, we use the definition.\[\begin{array}{rcl} \displaystyle \lim_{x \to \infty} \left[ f(x) - y \right] & = & \displaystyle \lim_{x \to \infty}\left( 3x - 1 + e^{-2x} - (3x - 1) \right) \\[6pt] & = & \displaystyle \lim_{x \to \infty} e^{-2x} \\[6pt] & = & 0 \end{array}\nonumber\]Thus, we have proven that \(y = 3x - 1\) is a slant asymptote of \(f(x) = 3x - 1 + e^{-2x}\) as \(x \to \infty\).

Domain: Since \(R(x)\) is a rational function and its denominator, \(x^2 + 1\), is always greater than or equal to \(1\), the domain is \(\mathbb{R} = (-\infty, \infty)\).

Intercepts: \((0,0)\) is the only intercept.

Symmetry: \(R(-x) = \frac{(-x)^2}{(-x)^2 + 1} = \frac{x^2}{x^2 + 1} = R(x)\). Therefore, \(R\) is an even function. As such, we will limit the rest of our investigation to the interval \([0, \infty)\) and reflect our graph across the \(y\)-axis when we are finished.

Vertical Asymptote(s): Since the domain of \(R\) is all real numbers and \(R\) is continuous, there is not going to be a vertical asymptote.

End Behavior:\[\begin{array}{rcl} \displaystyle \lim_{x \to \infty}{R(x)} & = & \displaystyle \lim_{x \to \infty}{\dfrac{x^2}{x^2 + 1}} \\[6pt] & = & \displaystyle \lim_{x \to \infty}{\dfrac{1}{1 + \frac{1}{x^2}}} \\[6pt] & = & \dfrac{1}{1 + 0} \\[6pt] & = & 1 \\[6pt] \end{array}\nonumber\]Thus, \(R\) has a horizontal asymptote of \(y = 1\) as \(x \to \infty\).

First Derivative Information:\[\begin{array}{rcl} R^{\prime}(x) & = & \dfrac{2x(x^2 + 1) - x^2(2x)}{(x^2 + 1)^2} \\[6pt] & = & \dfrac{2x(x^2 + 1 - x^2)}{(x^2 + 1)^2} \\[6pt] & = & \dfrac{2x}{(x^2 + 1)^2} \\[6pt] \end{array}\nonumber\]\(R^{\prime}(x) = 0\), when \(x = 0\). Otherwise, \(R^{\prime}(x)\) always exists. Thus, the only critical point of \(R\) is \((0,0)\).

For \(x \gt 0\), \(R^{\prime}(x) \gt 0\). Therefore, \(R\) is increasing on the interval \((0,\infty)\). Since \(R\) is an even function, \(R\) will be decreasing for \(x \lt 0\). This means that \((0,0)\) is a local minimum. All of the information can be seen on the number line for \(R^{\prime}\) in Figure \(\PageIndex{1A}\) below.

Second Derivative Information:\[\begin{array}{rcl} R^{\prime\prime}(x) & = & \dfrac{2(x^2 + 1)^2 - 2x \cdot 2(x^2 + 1) \cdot 2x)}{(x^2 + 1)^4} \\[6pt] & = & \dfrac{(x^2 + 1)\left(2(x^2 + 1) - 8 x^2 \right)}{(x^2 + 1)^4} \\[6pt] & = & \dfrac{-6x^2 + 2}{(x^2 + 1)^3} \\[6pt] & = & \dfrac{-2(3 x^2 - 1)}{(x^2 + 1)^3} \\[6pt] \end{array}\nonumber\]\(R^{\prime\prime}(x) = 0\), when \(x = \pm \frac{1}{\sqrt{3}}\). Otherwise, \(R^{\prime\prime}(x)\) always exists.

All of the information for \(R^{\prime\prime}\) can be seen on the number line for \(R^{\prime\prime}\) in Figure \(\PageIndex{1B}\) below (again, we only concern ourselves with \(x\)-values greater than \(0\) because \(R\) is even).

Graph Like a Boss: Before graphing, it's best to collect all the important information into one clean "pile." It's also required that you compute the function values for \(R\) wherever there are local extrema, inflection points, and intercepts.

Symmetry: Even

Horizontal Asymptote: \(y = 1\) as \(x \to \infty\)

Domain: We require \(1 - x^2 \ge 0 \implies 1 \ge x^2 \implies x \in [-1,1]\).

Intercepts: \((0,0)\) and \((\pm 1, 0)\) are the intercepts.

Symmetry: \(f(-x) = -x \sqrt{1 - (-x)^2} = -x \sqrt{1 - x^2} = -f(x)\). Therefore, \(f\) is an odd function. As such, we will limit the rest of our investigation to the interval \([0, 1]\) and reflect our graph about the origin when we are finished.

Vertical Asymptote(s): Since the domain of \(f\) is a closed interval and \(f\) is continuous, there is not going to be a vertical asymptote.

End Behavior: Since the domain of \(f\) does not extend to infinity (in either direction), there is no end behavior to speak of.

First Derivative Information:\[\begin{array}{rcl} f^{\prime}(x) & = & \sqrt{1 - x^2} + x \cdot \dfrac{- 2 x}{2 \sqrt{1 - x^2}} \\[6pt] & = & \sqrt{1 - x^2} - \dfrac{x^2}{\sqrt{1 - x^2}} \\[6pt] & = & \dfrac{1 - x^2 - x^2}{\sqrt{1 - x^2}} \\[6pt] & = & \dfrac{1 - 2 x^2}{\sqrt{1 - x^2}} \\[6pt] \end{array}\nonumber\]\(f^{\prime}(x) = 0\), when \(x = \pm \frac{1}{\sqrt{2}}\), and \(f^{\prime}(x)\) does not exist at \(x = \pm 1\). Since \(f\) is odd, we will only worry about \(x = \frac{1}{\sqrt{2}}\) and \(x = 1\). The corresponding critical points are \((1,0)\) and \(\left( \frac{1}{\sqrt{2}}, \frac{1}{2} \right)\). We first draw the number line.

Testing each interval (and only substituting these values into the numerator of \(f^{\prime}\) because the denominator is always positive), we get \(f^{\prime}\left(\frac{1}{2}\right) \gt 0\) and \(f^{\prime}\left( \frac{3}{4} \right) \lt 0\). Therefore, \(f\) is increasing on \(\left( 0, \frac{1}{\sqrt{2}}\right)\) and decreasing on \(\left( \frac{1}{\sqrt{2}}, 0 \right)\). Hence, \(f\) has a local maximum at the point \(\left( \frac{1}{\sqrt{2}}, \frac{1}{2} \right)\). All of this information can be seen on the number line for \(f^{\prime}\) in Figure \(\PageIndex{4A}\) below.

Second Derivative Information:\[\begin{array}{rcl} f^{\prime\prime}(x) & = & \dfrac{-4x \sqrt{1 - x^2} - \left(1 - 2x^2 \right) \cdot \dfrac{-x}{\sqrt{1 - x^2}}}{1 - x^2} \\[6pt] & = & \dfrac{-4x \left( 1 - x^2 \right) + x \left(1 - 2x^2 \right)}{\left(1 - x^2\right)^{3/2}} \\[6pt] & = & \dfrac{x \left(2 x^2 - 3 \right)}{\left(1 - x^2\right)^{3/2}} \\[6pt] \end{array}\nonumber\]\(f^{\prime\prime}(x) = 0\), when \(x = \pm \frac{\sqrt{6}}{2}\) and when \(x = 0\). Moreover, \(f^{\prime\prime}\) does not exist when \(x = \pm 1\). Again, since \(f\) is odd, we will only concern ourselves with \(x = 0\), \(x = \frac{\sqrt{6}}{2}\), and \(x = 1\); however, \(\frac{\sqrt{6}}{2} \gt 1\) and, as such, is outside of the domain of \(f\). Hence, we only need to worry about \(x = 0\) and \(x = 1\). Since \(f^{\prime\prime}\left( \frac{1}{2} \right) \lt 0\), \(f\) is concave down on \((0,1)\) (see Figure \(\PageIndex{4B}\)).

Graph Like a Boss:

Symmetry: Odd

Domain: The domain of \(f\) is the set of all real numbers \(x\) except \(x=1\).

Intercepts: We can see that when \(x=0, \,f(x)=0\), so \((0,0)\) is the only intercept.

Symmetry: Since \(f(-x) = \frac{(-x)^2}{(-x) - 1} = \frac{x^2}{-x - 1} = - \frac{x^2}{x + 1} \neq \pm f(x)\), \(f\) is neither odd nor even.

Vertical Asymptote(s): To check for vertical asymptotes, look at where the denominator is zero. Here, the denominator is zero at \(x=1\). Looking at both one-sided limits as \(x \to 1\), we find\[\displaystyle \lim_{x \to 1^+}\dfrac{x^2}{x-1}=\infty\nonumber\]and\[\displaystyle \lim_{x \to 1^-}\dfrac{x^2}{x-1}=-\infty.\nonumber\]Therefore, \(x=1\) is a vertical asymptote, and we have determined the behavior of \(f\) as \(x\) approaches \(1\) from the right and the left.

End Behavior: Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, \(f\) must have a slant asymptote. To find the slant asymptote, use long division of polynomials to write\[f(x)=\dfrac{x^2}{x-1}=x+1+\dfrac{1}{x-1}.\nonumber\]Since \(\frac{1}{x-1} \to 0\) as \(x \to \pm \infty, f(x)\) approaches the line \(y=x+1\) as \(x \to \pm \infty\). The line \(y=x+1\) is a slant asymptote for \(f\).

First Derivative Information: Calculate the first derivative:\[f^{\prime}(x)=\dfrac{(x-1)(2x)-x^2(1)}{(x-1)^2}=\dfrac{x^2-2x}{(x-1)^2}.\nonumber\]We have \(f^{\prime}(x)=0\) when \(x^2-2x=x(x-2)=0\). Therefore, \(x=0\) and \(x=2\) are critical points. Since \(f\) is undefined at \(x=1\), we need to divide the interval \((-\infty,\infty)\) into the smaller intervals \((-\infty,0), (0,1), (1,2)\), and \((2,\infty)\), and choose a test point from each interval to evaluate the sign of \(f^{\prime}(x)\) in each of these smaller intervals. Doing so, we get the following sign chart for \(f^{\prime}\).

From this sign chart, we see that \(f\) has a local maximum at \(x=0\) and a local minimum at \(x=2\). The value of \(f\) at the local maximum is \(f(0)=0\) and the value of \(f\) at the local minimum is \(f(2)=4\). Therefore, \((0,0)\) and \((2,4)\) are important points on the graph.

Second Derivative Information: Calculate the second derivative:\[\begin{array}{rcl} f^{\prime\prime}(x) & = & \dfrac{(x-1)^2(2x-2)-2(x-1)(x^2-2x)}{(x-1)^4} \\[6pt] & = & \dfrac{2(x-1)[(x-1)^2-(x^2-2x)]}{(x-1)^4}\\[6pt] & = & \dfrac{2[x^2-2x+1-x^2+2x]}{(x-1)^3}\\[6pt] & = & \dfrac{2}{(x-1)^3}. \\[6pt] \end{array}\nonumber\]We see that \(f^{\prime\prime}(x)\) is never zero or undefined for \(x\) in the domain of \(f\). Since \(f\) is undefined at \(x=1\), to check concavity we just divide the interval \((-\infty,\infty)\) into the two smaller intervals \((-\infty,1)\) and \((1,\infty)\), and choose a test point from each interval to evaluate the sign of \(f^{\prime\prime}(x)\) in each of these intervals.

Graph Like a Boss: When graphing this function by hand, it's best first to plot all the key points (\(\left( 0, 0 \right)\) and \(\left( 2, 4 \right)\)) as well as the slant asymptote. Once that is done, you can get a good idea of the scale of things. We arrive at the following graph for \(f\) from the information gathered.

Domain: Since the cube-root function is defined for all real numbers \(x\) and \((x-1)^{2/3}=(\sqrt[3]{x-1})^2\), the domain of \(f\) is all real numbers.

Intercepts: To find the \(y\)-intercept, evaluate \(f(0)\). Since \(f(0)=1\), the \(y\)-intercept is \((0,1)\). To find the \(x\)-intercept, solve \((x-1)^{2/3}=0\). The solution of this equation is \(x=1\), so the \(x\)-intercept is \((1,0)\).

Symmetry: Since\[\begin{array}{rcl} f(-x) & = & (-x - 1)^{2/3} \\[6pt] & = & (x + 1)^{2/3} \\[6pt] & \neq & \pm f(x), \\[6pt] \end{array}\nonumber\]we know that \(f\) is neither odd nor even.

Vertical Asymptote(s): The function has no vertical asymptotes.

End Behavior: Since \(\displaystyle \lim_{x \to \pm \infty}(x-1)^{2/3}=\infty\), the function continues to grow without bound as \(x \to \infty\) and \(x \to -\infty\).

First Derivative Information: To determine where \(f\) is increasing or decreasing, calculate \(f^{\prime}\). We find\[f^{\prime}(x)=\dfrac{2}{3}(x-1)^{-1/3}=\dfrac{2}{3(x-1)^{1/3}}\nonumber\]This function is not zero anywhere, but it is undefined when \(x=1\). Therefore, the only critical point is \(x=1\). Divide the interval \((-\infty,\infty)\) into the smaller intervals \((-\infty,1)\) and \((1,\infty)\), and choose test points in each of these intervals to determine the sign of \(f^{\prime}(x)\) in each of these smaller intervals.

We conclude that \(f\) has a local minimum at \(x=1\). Evaluating \(f\) at \(x=1\), we find that the value of \(f\) at the local minimum is zero. Note that \(f^{\prime}(1)\) is undefined, so to determine the behavior of the function at this critical point, we need to examine \(\displaystyle \lim_{x \to 1} f^{\prime}(x)\). Looking at the one-sided limits, we have\[\displaystyle \lim_{x \to 1^+}\dfrac{2}{3(x-1)^{1/3}}=\infty \text{ and } \displaystyle \lim_{x \to 1^-}\dfrac{2}{3(x-1)^{1/3}} = -\infty.\nonumber\]Therefore, \(f\) has a cusp at \(x=1\).

Second Derivative Information: To determine concavity, we calculate the second derivative of \(f:\)\[f^{\prime\prime}(x)=-\dfrac{2}{9}(x-1)^{-4/3}=\dfrac{-2}{9(x-1)^{4/3}}.\nonumber\]We find that \(f^{\prime\prime}(x)\) is defined for all \(x\), but is undefined when \(x=1\). Therefore, divide the interval \((-\infty,\infty)\) into the smaller intervals \((-\infty,1)\) and \((1,\infty)\), and choose test points to evaluate the sign of \(f^{\prime\prime}(x)\) in each of these intervals.

We conclude that \(f\) is concave down everywhere.

Graph Like a Boss:

Combining all this information, we arrive at the following graph for \(f\).