10.3: Inverse Functions and Logarithms

Learning Objectives

1. Explain the concept of inverse function from both algebraic and geometric points of view: given a function, determine whether (and for what restricted domain) an inverse function can be defined and sketch that inverse function.
2. Describe the relationship between the domain and range of a function and the range and domain of its inverse function. (Review Appendix C.5).
3. Apply these ideas to the logarithm, which is the inverse of an exponential function.
4. Reproduce the calculation of the derivative of \(\ln (x)\) using implicit differentiation.

In this chapter we defined the new function \(e^{x}\) and computed its derivative. Paired with this newcomer is an inverse function, the natural logarithm, \(\ln (x)\). Recall the following key ideas:

Given a function \(y=f(x)\), its inverse function, denoted \(f^{-1}(x)\) satisfies

\[f\left(f^{-1}(x)\right)=x, \quad \text { and } \quad f^{-1}(f(x))=x . \nonumber \]

• The range of \(f(x)\) is the domain of \(f^{-1}(x)\) (and vice versa), which implies that in many cases, the relationship holds only on some subset of the original domains of the functions.
• The functions \(f(x)=x^{n}\) and \(g(x)=x^{1 / n}\) are inverses of one another for all \(x\) when \(n\) is odd.
• The domain of a function (such as \(y=x^{2}\) or other even powers) must be restricted (e.g. to \(x \geq 0\) ) so that its inverse function \((y=\sqrt{x})\) is defined.
• On that restricted domain, the graphs of \(f\) and \(f^{-1}\) are mirror images of one another about the line \(y=x\). Essentially, this stems from the fact that the roles of \(x\) and \(y\) are interchanged.

Mastered Material Check

19. Are \(f(x)=x^{n}\) and \(g(x)=x^{1 / n}\) also inverses of one another for even integer \(n\) ? Is this true for all \(x\) ?
20. What is the inverse function for \(y=x ?\) Over what range of values is the inverse defined?
21. What is the inverse function to \(y=x^{2 / 3}\) and over what domain are the two functions inverses of one another?

The natural logarithm is an inverse function for \(e^{x}\)

Figure 10.6: The function \(y=e^{x}\) is shown with its inverse, \(y=\ln x\).

For \(y=f(x)=e^{x}\) we define an inverse function, shown on Figure 10.6. We call this function the logarithm (base \(e\) ), and write it as

\[y=f^{-1}(x)=\ln (x) \nonumber \]

We have the following connection: \(y=e^{x}\) implies \(x=\ln (y)\). The fact that the functions are inverses also implies that

\[e^{\ln (x)}=x \quad \text { and } \quad \ln \left(e^{x}\right)=x . \nonumber \]

The domain of \(e^{x}\) is \(-\infty<x<\infty\), and its range is \(x>0\). For the inverse function, this domain and range are interchanged, meaning that \(\ln (x)\) is only defined for \(x>0\) (its domain) and returns values in \(-\infty<x<\infty\) (its range). As shown in Figure 10.6, the functions \(e^{x}\) and \(\ln (x)\) are reflections of one another about the line \(y=x\).

Note symmetry about the line \(y=x\) for this graph of \(f(x)=x^{n}\) and \(g(x)=x^{1 / n}\). Adjust the slider for \(n\) to see how even and odd powers behave. What do you notice about the domain over which \(g(x)\) is defined? Adjust the slider for \(a\) to observe "corresponding points" on the two graphs.

Properties of the logarithm stem directly from properties of the exponential function. A review of these is provided in Appendix B.2. Briefly,

1. \(\ln (a b)=\ln (a)+\ln (b)\)
2. \(\ln \left(a^{b}\right)=b \ln (a)\)
3. \(\ln (1 / a)=\ln \left(a^{-1}\right)=-\ln (a)\)

Mastered Material Check

22. Give algebraic justification of the three properties of logarithms.

Featured Problem 10.2 (Agroforestry): In agroforestry, the farming of crops is integrated with growing of trees to benefit productivity and maintain the health of an ecosystem. A tree can provide advantage to nearby plants by creating better soil permeability, higher water retention, and more stable temperatures. At the same time, trees produce shade and increased competition for nutrients. Both the advantage \(A(x)\) and the shading \(S(x)\) depend on distance from the tree, with shading a dominant negative effect right under the tree.

Figure 10.7: Too close to a tree, shading (grey) \(S(x)\) interferes with crop growth. Just beyond this region, the advantage \(A(x)\) to crop growth outweighs any disadvantage due to shading. We seek to find the optimal distance \(x\) for planting the crops.

Suppose that at a distance \(x\) from a given tree species, the net benefit \(B\) to a crop plant can be expressed as the difference

\[B(x)=A(x)-S(x), \quad \text { where } A(x)=\alpha e^{-x^{2} / a^{2}}, \quad S(x)=\beta e^{-x^{2} / b^{2}}, \alpha \beta, a, b>0 \nonumber \]

1. How far away from the tree will the two influences break even?
2. Find the optimal distance to plan crops so that they derive maximal benefit from the nearby tree.

The Advantage \(A(x)\), the shading effect \(S(x)\), and the net benefit \(B(x)\) for a crop as functions of distance \(x\) from a tree are shown here. Move the sliders to see how the spatial range \(a\) and the magnitude \(\beta\) affect the graphs. Tangent lines to the graphs of \(y=e^{x}\) and \(y=\ln (x)\) at corresponding points are mirror images about the line \(y=x\). Adjust the slider to see the tangent lines at various points along the curves. What do we mean by "corresponding points"?

Derivative of \(\ln (x)\) by implicit differentiation

Implicit differentiation is helpful whenever an inverse function appears. Knowing the derivative of the original function allows us to compute the derivative of its inverse by using their relationship. We use implicit differentiation to find the derivative of \(y=\ln (x)\).

First, restate the relationship in the inverse form, but consider \(y\) as the dependent variable - that is think of \(y\) as a quantity that depends on \(x\) :

\[y=\ln (x) \quad \Rightarrow \quad e^{y}=x \quad \Rightarrow \quad \frac{d}{d x} e^{y(x)}=\frac{d}{d x} x . \nonumber \]

Applying the chain rule to the left hand side,

\[\frac{d e^{y}}{d y} \frac{d y}{d x}=1 \quad \Rightarrow \quad e^{y} \frac{d y}{d x}=1 \quad \Rightarrow \quad \frac{d y}{d x}=\frac{1}{e^{y}}=\frac{1}{x} . \nonumber \]

We have thus shown the following:

The derivative of \(\ln (x)\) is \(1 / x\) :

\[\frac{d \ln (x)}{d x}=\frac{1}{x} . \nonumber \]

Inverse functions are mirror images of one another about the line \(y=x\), since the role of independent and dependent variables are switched. Their tangent lines are also mirror images about the same line.

Tangent lines to the graphs of \(y = e^x\) and \(y=\ln (x)\) at corresponding points are mirror images about the line \(y = x\). Adjust the slider to see the tangent lines at various points along the curves. What the role of independent and dependent variables are switched. Their tangent do we mean by “corresponding points”?