Problem 69 - Essential Calculus Early Transcendentals

Question

(a) Find the domain of \(f(x)=\ln \left(e^{x}-3\right)\) (b) Find \(f^{-1}\) and its domain.

Step-by-Step Solution

Answer

(a) Domain of \( f(x) \): \( (\ln(3), \infty) \); (b) \( f^{-1}(x) = \ln(e^x + 3) \), domain \((-\infty, \infty)\).

Step 1: Understand the Function Domain Restrictions

To find the domain of the function \( f(x) = \ln(e^x - 3) \), we need to ensure that the expression inside the natural logarithm, \( e^x - 3 \), is positive. This is because the logarithm is only defined for positive numbers. Thus, we require \( e^x - 3 > 0 \).

Step 2: Solve the Inequality \(e^x - 3 > 0\)

Rearrange the inequality: \[ e^x > 3 \] Taking the natural logarithm of both sides, we get \[ \ln(e^x) > \ln(3) \] Since \( \ln(e^x) = x \), it follows that \[ x > \ln(3) \].

Step 3: State the Domain of \( f(x) \)

The domain of \( f(x) = \ln(e^x - 3) \) is all real numbers greater than \( \ln(3) \). In interval notation, the domain is \( (\ln(3), \infty) \).

Step 4: Express the Function as a One-to-One Mapping for \(f^{-1}(x)\)

To find the inverse function \( f^{-1} \), first express \( y = \ln(e^x - 3) \). Solving for \( x \) involves exponentiating both sides: \[ e^y = e^x - 3 \] Rearrange to solve for \( x \): \[ e^x = e^y + 3 \] Taking the natural logarithm gives: \[ x = \ln(e^y + 3) \].

Step 5: State the Inverse Function \( f^{-1}(x) \)

The inverse function \( f^{-1}(x) \) is \( \ln(e^x + 3) \).

Step 6: Determine the Domain of \( f^{-1}(x) \)

The expression inside the logarithm for \( f^{-1}(x) = \ln(e^x + 3) \) must be positive. Since \( e^x + 3 \) is always positive for all real numbers \( x \), the domain of \( f^{-1}(x) \) is all real numbers, \( (-\infty, \infty) \).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial in calculus because it specifies all the possible input values (x-values) for which the function is defined. For the function given, \( f(x) = \ln(e^x - 3) \), the critical point is the expression inside the logarithm function, \( e^x - 3 \). The natural logarithm function, \( \ln(x) \), only accepts positive numbers as its input. Hence, the condition \( e^x - 3 > 0 \) must be met for the function to exist.

• Firstly, rearrange the inequality to \( e^x > 3 \).
• Secondly, take the natural logarithm of both sides to solve for \( x \).

Since \( \ln(e^x) = x \), the solution is \( x > \ln(3) \). Therefore, the domain of this function is all real numbers greater than \( \ln(3) \), presented in interval notation as \( (\ln(3), \infty) \). This means any x-value within this range will keep \( e^x - 3 \) positive, ensuring the function remains defined.

Inverse Functions

An inverse function essentially reverses the effect of a given function. For the function \( f(x) = \ln(e^x - 3) \), the task is to find its inverse \( f^{-1}(x) \). When creating an inverse, swap the roles of \( x \) and \( y \) in the function equation, and solve for the new value of \( x \):

• Start with \( y = \ln(e^x - 3) \).
• To isolate \( x \), exponentiate both sides to get \( e^y = e^x - 3 \).

From here, rearrange the equation to \( e^x = e^y + 3 \). Taking the logarithm of both sides results in the expression \( x = \ln(e^y + 3) \), thus defining the inverse function as \( f^{-1}(x) = \ln(e^x + 3) \).The inverse function allows you to reverse the result of the original function back to the input value, making it a valuable tool in calculus.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a fundamental function in calculus that serves as the inverse of the exponential function \( e^x \). This logarithmic function only acts over positive real numbers; hence, the input for \( \ln(x) \) must be positive, guiding us to decide the domain of functions like \( f(x) = \ln(e^x - 3) \).

• The expression \( \ln(e^x) \) simplifies directly to \( x \).
• It implies that applying both \( e^x \) and \( \ln(x) \) to a value essentially returns the value itself, given the input is positive.

Additionally, the natural logarithm has important properties in simplifying expressions and solving exponential equations due to its unique relationship with the base \( e \), approximately equal to 2.718. These characteristics make it indispensable in calculus, particularly in solving problems concerning growth and decay, just like determining function domains and calculating inverses.