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Functions and single-variable calculus A course resource for Math 218

Ellen Goldstein, Aaron Greicius

Section 2.14 Exponential and logarithmic curve sketching

Subsection Curve sketching

The same techniques for curve sketching developed in Section 2.6 can be applied to functions involving exponential and logarithmic expressions. Recall that a complete curve sketch analysis of a function will result in a graph that reflects the various properties enumerated in Procedure 2.6.1.
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Example 2.14.1. Logarithmic curve sketching.

Provide a graph of \(f(x)=\ln(x^5-32)\) that includes all the details listed in Procedure 2.6.1.
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Solution.
We first compute the domain \(D\) of \(f\text{.}\) Since \(\ln t\) is defined if and only if \(t>0 \text{,}\) \(D=\{x\in \R\mid x^5-32> 0\}\text{.}\) We solve:
\begin{align*} x^5-32 \amp > 0\\ x^5 \amp > 32\\ x \amp > 32^{1/5}\\ x \amp > 2\text{.} \end{align*}
Here we have used the fact that the function \(g(x)=x^5\) is an increasing function on its domain, and has inverse \(g^{-1}(x)=x^{1/5}\text{.}\) (We could also have solved the inequality using our trusty friend the sign diagram.) We conclude that \(D=(2,\infty)\text{.}\) Note also, that our limit analysis, and the fact that \(f\) is continuous, implies that the range of \(f\) is \(\R\text{.}\)
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Since \(0\notin D\text{,}\) \(f\) has no \(y\)-intercept. For the \(x\)-intercept(s), we solve
\begin{align*} f(x) \amp =0\\ \ln(x^5-32) \amp = 0\\ x^5-32\amp = e^0\\ x^5-32 \amp = 1\\ x^5 \amp = 33 \\ x \amp =33^{1/5}=\sqrt[5]{33}\text{.} \end{align*}
Thus \((33^{1/5},0)\) is the \(x\)-intercept of the graph of \(f\text{.}\)
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Since \(D=(2,\infty)\text{,}\) for the end behavior of \(f\) we compute
\begin{align*} \lim\limits_{x\to 2}f(x) \amp = \lim\limits_{x\to 2^+}\ln(x^2-32) \\ \amp = \lim\limits_{t\to 0^+}\ln t \amp (x\to 2^+\implies x^5-32\to 0^+)\\ \amp = -\infty \\ \lim\limits_{x\to \infty}\ln(x^5-32) \amp =\infty \text{,} \end{align*}
where the last limit follows since \(x^5-32\to \infty\) as \(x\to infty\text{,}\) and thus \(\ln(x^5-32)\to \infty\) as well. We conclude that the graph of \(f\) has a vertical asymptote at \(x=2\) and no horizontal asymptote.
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We now determine the critical points of \(f\text{.}\) We have
\begin{align*} f'(x) \amp =\frac{5x^4}{x^5-32}\text{,} \end{align*}
which is defined for all \(x\in D\) and never equal to zero on \(D\text{,}\) since \(0\notin D\text{.}\) Indeed, since \(x^4\) and \(x^5-32\) are positive for all \(x\in D\text{,}\) we see that \(f'\) is positive everywhere, making \(f\) an increasing function on its domain with no critical points. We have thus also determined the intervals of monotonicity of \(f\text{.}\)
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Lastly, we examine the concavity of \(f\) and possible inflection points. We have
\begin{align*} f''(x) \amp = \frac{20x^3(x^5-32)-5x^4(5x^4)}{(x^5-32)^2} \\ \amp = \frac{20x^8-640x^3-25x^8}{(x^5-32)^2} \\ \amp = \frac{-x^3(5x^5+640)}{(x^5-32)^2} \text{.} \end{align*}
It is easy to see that the numerator above is negative for all \(x\in D\text{,}\) and thus that \(f\) is concave down throughout its domain. As a result, \(f\) has no inflection points. The graph in Figure 2.14.2 nicely summarizes our findings.
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Example 2.14.3. Surge function.

The function \(f(x)=xe^{-x}\text{,}\) with domain \(D=[0,\infty)\) is an example of a surge function.
  1. Find the \(x\)- and \(y\)-intercepts of the graph of \(f\text{.}\)
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  2. You may take for granted that \(\lim\limits_{x\to \infty}f(x)=0\text{.}\) We will develop tools for computing this limit in Section 2.15. Intuitively the idea is that as \(x\to \infty\text{,}\) the term \(e^{-x}\) approaches zero much faster than the term \(x\) approaches \(\infty\text{.}\)
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  3. Find and classify all critical points of \(f\text{,}\) determine intervals of monotonicity, and identify any absolute extrema of \(f\) using Procedure 2.7.7.
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  4. Find the intervals of constant concavity of \(f\) and identify any inflection points.
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Solution.
We have \(f(0)=0\) and \(f(x)=0\) if and only if \(xe^{-x}=0\text{,}\) if and only if \(x=0\) (since \(e^{-x}> 0\) for all \(x\)). Thus \((0,0)\) is the the sole \(x\)- and \(y\)-intercept of the graph of \(f\text{.}\)
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For the rest of our analysis, we first need to compute \(f'\) and \(f''\text{:}\)
\begin{align*} f'(x) \amp = e^{-x}-xe^{-x}=(1-x)e^{-x}\\ f''(x)\amp = -e^{-x}-(1-x)e^{-x}=(x-2)e^{-x} \text{.} \end{align*}
From these formulas it follows easily that \(x=1\) is the only critical point of \(f\text{,}\) that \(f'(x)> 0\) if and only \(x< 1\text{,}\) and \(f'(x)< 0\) if and only if \(x> 1\text{,}\) as summarized by the sign diagram below.
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Sign diagram for derivative of f
From the sign diagram, we conclude that \(f(1)=e^{-1}=1/e\) is a local maximum value of \(f\text{.}\) Comparing \(f(1)=1/e\) to \(f(0)=0\) and \(\lim\limits_{x\to \infty}f(x)=0\text{,}\) we conclude from Procedure 2.7.7 that \(f(1)=1/e\) is in fact the absolute maximum value of \(f\) on \(D\text{.}\)
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Moving now to concavity, the properties of \(f''\) are summarized by the corresponding sign diagram below.
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Sign diagram for second derivative of f
From this we see that \(f\) is concave down on \((0,2)\) and concave up on \((2,\infty)\text{,}\) and hence that the point \(P=(2,f(2))=(2, 2/e^2)\) is an inflection point of the graph of \(f\text{.}\) The graph in Figure 2.14.4 reflects the various details of \(f\) our analysis has uncovered.
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Graph of surge function
Figure 2.14.4. Graph of the surge function \(f(x)=xe^{-x}\)
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For our last example we will consider a general logistic growth function
\begin{align*} f(x) \amp =\frac{A}{1+ce^{-kx}}\text{,} \end{align*}
where \(A\text{,}\) \(c\text{,}\) and \(k\) are fixed positive constants. As we wil discover in Example 2.14.5, starting from \(x=0\text{,}\) the values \(f(x)\) of a logistic growth begins growing very rapidly at first, but then level off as they approach a fixed ceiling valhue. Because of this trait, whereas exponential functions \(f(x)=Ae^{kx}\) are used to model unconstrained growth, logistic functions are often used to model growth that is constrained by some factor: e.g., a population whose growth is constrained by a limited amount of food resources.
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Example 2.14.5. Logistic growth.

A function of the form
\begin{align} f(x) \amp =\frac{A}{1+ce^{-kx}}\text{,}\tag{2.53} \end{align}
where \(A\text{,}\) \(c\text{,}\) and \(k\) are fixed positive constants, is called a logistic function (or logistic growth function).
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Perform a complete curve sketching analysis of \(f\) on the domain \(D=[0,\infty)\text{,}\) with the additional assumption that \(c> 1\text{.}\) Include all details enumerated in Procedure 2.6.1 and identify any any absolute extrema. Some of your details will be expressed in terms of the constants \(A,c,k\text{.}\)
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Solution.
Since the numerator and denominator in the formula (2.53) are both positive, \(f\) has no zeros on its domain, and hence there is no \(x\)-intercept. Since \(f(0)=A/(1+ce^0)\text{,}\) the \(y\)-intercept of the graph of \(f\) is \((0,A/(1+c)\text{.}\)
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The value \(f(0)=A/(1+c)\) determines the left end behavior of \(f\text{.}\) For the right end behavior, we compute the limit
\begin{align*} \lim\limits_{x\to \infty}f(x) \amp = \lim\limits_{x\to \infty}\frac{A}{1+ce^{-kx}}\\ \amp =\frac{A}{1+c\lim\limits_{x\to \infty}e^{-kx}}\\ \amp =\frac{A}{1+0}\\ \amp =A\text{.} \end{align*}
Thus the graph of \(f\) has the horiztontal asymptote \(y=A\text{.}\)
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Continuing with our analysis, we now look at the first and second derivatives of \(f\text{:}\)
\begin{align*} f'(x) \amp =-A(1+ce^{-kx})^{-2}(1+ce^{-kx})'\\ \amp = \frac{ckAe^{-kx}}{(1+ce^{-kx})^2}\\ f''(x) \amp = ckA\frac{-ke^{-kx}(1+ce^{-kx})^2-2e^{-kx}(1+ce^{-kx})(-cke^{-kx})}{(1+ce^{-kx})^4}\\ \amp = \frac{ck^2Ae^{-kx}(1+ce^{-kx})(ce^{-kx}-1)}{(1+ce^{-kx})^4}\text{.} \end{align*}
From since the numerator and denominator for \(f'\) are both easily seen to be positive for all \(x\in D\text{,}\) we see that \(f\) is increasing on its entire domain, and has no critical points.
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Moving to \(f''\) we see that the denominator, as well as all of the terms in the numerator, with the exception of \((ce^{-kx}-1)\) are positive. Thus the sign of \(f''\) is equal to the sign of \(g(x)=ce^{-kx}-1\text{.}\) Performing a sign diagram for \(g\text{,}\) we first solve
\begin{align*} g(x) \amp = 0\\ ce^{-kx} \amp =1\\ e^{-kx} \amp =\frac{1}{c}\\ -kx \amp = \ln(1/c)=-\ln c\\ x \amp = \ln(c)/k \text{.} \end{align*}
Note that since we assume \(c> 1\text{,}\) we have \(\ln c> 0\text{,}\) whence \(x=\ln(c)/k> 0\text{.}\) Since \(g(0)=c-1> 0\) and \(g(x)\to -1 \) as \(x\to \infty\text{,}\) we conclude that \(f''\) has the following sign diagram.
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Sign diagram for second derivative of logistic function
We conclude that \(f''\) is concave up on \((0,\ln(c)/k)\) and concave down on \((\ln(c)/k,\infty)\text{,}\) and thus has an inflection point at
\begin{align*} P= \amp (\ln(c)/k, f(\ln(c)/k))\\ \amp =(\ln(c)/k, A/(1+ce^{-\ln c}))\\ \amp =(\ln(c)/k, A/2)\text{.} \end{align*}
Interestingly, we see that the unique inflection point of the graph of \(f\) occurs at the unique point where the \(y\)-coordinate is \(A/2\text{,}\) exactly half of the ceiling value of the function!
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The graph in Figure 2.14.6 assembles all of the details of our analysis.
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Graph of logistic growth function
Figure 2.14.6. Graph of \(f(x)=\frac{A}{1+ce^{-kx}}\text{,}\) \(A,k> 0\text{,}\) \(c> 1\)
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