Problem 32 - Calculus of a Single Variable

Question

Inverse Functions In Exercises \(29-32\) , illustrate that the functions are inverses of each other by graphing both functions on the same set of coordinate axes. $$ \begin{array}{l}{f(x)=e^{x-1}} \\ {g(x)=1+\ln x}\end{array} $$

Step-by-Step Solution

Answer

Yes, the functions \(f(x)=e^{x-1}\) and \(g(x)=1+ \ln(x)\) are inverses of each other. This was proven both algebraically and confirmed graphically.

Step 1: Define the functions

The functions given are \(f(x)=e^{x-1}\) and \(g(x)=1+ \ln(x)\)

Step 2: Show algebraically they are inverse functions

To show two functions are inverses of each other, we need to prove:- \(f(g(x))=x\) which means if we substitute \(g(x)\) in place of x in function \(f\), the output should be x. - Similarly, \(g(f(x))=x\) which means if we substitute \(f(x)\) in place of x in function \(g\), the output should be \(x\). Let's do it:Showing \(f(g(x))=x\): Substitute \(g(x) = 1+ \ln(x)\) in \(f(x)=e^{x-1}\), we get:\(f(g(x)) = e^{(1+ \ln(x) -1)} = e^{\ln(x)}=x\). So, \(f(g(x))=x\).Showing \(g(f(x))=x\): Substitute \(f(x) = e^{x-1}\) in g\(x) = 1+ \ln(x)\), we get:\(g(f(x)) = 1+ \ln(e^{x-1}) = 1+ (x-1) = x\). So, \(g(f(x))=x\). Thus, the given functions are inverses of each other algebraically.

Step 3: Graph the functions

After proving algebraically, you can use any graphing software to plot the functions and see that they are reflections over the line y=x, further confirming they are inverses. The function \(f(x)=e^{x-1}\) is an exponential function shifting one unit to the right. The function \(g(x)=1+ \ln(x)\) is a logarithmic function shifting one unit up.

Key Concepts

Exponential and Logarithmic Functions

Exponential functions represent continuous growth or decay and are characterized by the equation of the form \( f(x) = a e^{kx} \), where \( a \) is a constant, \( e \) is the base of natural logarithms, and \( k \) is the rate of growth or decay. In the exercise, \( f(x) = e^{x-1} \) is an exponential function with base \( e \) and it is translated one unit to the right.

On the other hand, logarithmic functions are the inverses of exponential functions and are of the form \( g(x) = \text{log}_a(x) \) or, if using the natural logarithm, \( g(x) = \text{ln}(x) \). The function \( g(x) = 1+ \text{ln}(x) \) is a logarithmic function translated one unit upward. Logarithmic and exponential functions have unique properties that make them inverses of each other. For instance, the logarithm of a number is the exponent to which the base must be raised to produce that number:

• \( e^{\text{ln}(x)} = x \)
• \( \text{ln}(e^x) = x \)

When dealing with these functions, remember that they transform in opposite directions; as an exponential function increases rapidly, the logarithmic function rises slowly and vice versa. This concept is essential when working with inverse functions, as we aim to find a pair of functions where one 'undoes' the operation of the other.

Algebraic Proof of Inverse Functions

Proving that two functions are inverses algebraically involves demonstrating that each function, when composed with the other, returns the original input value. In the context of algebra, this approach confirms that operations of the functions cancel each other out, so if \( f(x) \) and \( g(x) \) are inverse functions, then:\( f(g(x)) = x \) and \( g(f(x)) = x \).

As seen in the solution provided, this proof is conducted by substituting one function into the other and simplifying to show that the result equals the identity function, \( x \). This is a rigorous method to ensure that the functions indeed reverse each other's actions:

• To prove \( f(g(x)) = x \), we replace \( g(x) \) within \( f(x) \) and simplify.
• To prove \( g(f(x)) = x \), we replace \( f(x) \) within \( g(x) \) and simplify.

This process is akin to verifying a lock and its key. Just as a key is made to fit perfectly into its corresponding lock, an inverse function is designed to fit perfectly into its original function to unlock the value \( x \).

Graphing Inverse Functions

Graphically, inverse functions can be confirmed by their symmetry with respect to the line \( y = x \). When plotted on the same set of axes, an original function and its inverse will appear as mirror images across this line. In practice, this means that for each point \( (a, b) \) on the graph of the original function, the inverse function will have a corresponding point \( (b, a) \).

To visualize this, you can employ a graphing tool or draw by hand. For the exponential function \( f(x) = e^{x-1} \), you would plot a curve that continuously rises from left to right. In contrast, the graph of its inverse, \( g(x) = 1 + \text{ln}(x) \), would rise from bottom to top – starting very steep and gradually flattening out.

By graphing both \( f(x) \) and \( g(x) \) on the same axes, you can confirm that they are inverses not only through algebraic proof but also visually, as they will form a symmetrical pattern across the line \( y = x \). This symmetry is a powerful visual cue to understand how inverse functions operate, further reinforcing the understanding that each function undoes the effect of the other.