Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(F(f(t), g(t))\) if \(F(x, y)=e^{x}+y^{2}\) and \(f(t)=\ln t^{2}\), \(g(t)=e^{t / 2} .\)

Step-by-Step Solution

Answer

\(t^2 + e^t\)

Step 1: Identify Functions to Substitute

The given problem involves functions: \(F(x, y) = e^x + y^2\), \(f(t) = \ln t^2\), and \(g(t) = e^{t/2}\). We need to substitute \(x\) and \(y\) in the function \(F(x, y)\) with \(f(t)\) and \(g(t)\) respectively.

Step 2: Substitute f(t) and g(t) into F(x,y)

Replace \(x\) with \(f(t) = \ln t^2\) and \(y\) with \(g(t) = e^{t/2}\). This gives us \(F(f(t), g(t)) = e^{\ln t^2} + (e^{t/2})^2\).

Step 3: Simplify Exponential Expressions

The expression \(e^{\ln t^2}\) simplifies to \(t^2\) because \(e^{\ln a} = a\). Also, \((e^{t/2})^2 = e^{t}\).

Step 4: Combine Simplified Results

Now that we have simplified the components, we combine them to find \(F(f(t), g(t)) = t^2 + e^t\). So the final expression for \(F(f(t), g(t))\) is \(t^2 + e^t\).

Key Concepts

Function Composition

Function composition is a powerful tool that allows you to build complex functions by combining simpler ones. Essentially, if you have two functions, say \( f(x) \) and \( g(x) \), composing them involves plugging the output of one function directly into the input of the other.
In the context of our exercise, function composition is used to replace variables within a function using other functions. To solve the problem, we needed to substitute \( x \) with \( f(t) = \ln t^2 \) and \( y \) with \( g(t) = e^{t/2} \) in the function \( F(x, y) = e^x + y^2 \).
This step-by-step substitution is vital in simplifying and solving the given expressions. Keep in mind:

• Identify the functions that need to be substituted.
• Carefully replace each variable.
• Ensure that the composed function remains within the scope of its domain.

Through function composition, we changed \( F(x, y) \) into \( F(f(t), g(t)) \), leading us to simplify it further.

Exponential Functions

Exponential functions are a key component of calculus and are characterized by a constant base raised to a variable exponent, like \( e^x \). Understanding how to manipulate these functions enables you to solve a variety of mathematical problems, particularly those involving growth and decay.
In solving the provided exercise, we encountered exponential functions when substituting into \( F(x, y) \). Our goal was to simplify exponential expressions like \( e^{\ln t^2} \) and \( (e^{t/2})^2 \):

• \( e^{\ln t^2} = t^2 \) because \( e^{\ln a} = a \) due to the inverse nature of exponential and logarithmic functions.
• \( (e^{t/2})^2 = e^t \), which follows from the property that \((a^m)^n = a^{m \cdot n}\).

This simplification is essential, as it transforms otherwise complicated expressions into more manageable forms. Understanding these fundamental properties of exponential functions can significantly aid in tackling similar problems effectively.

Natural Logarithm Properties

The natural logarithm \( \ln x \) is the inverse of the exponential function \( e^x \). It plays a crucial role in calculus by simplifying complex multiplication and division into addition and subtraction, respectively.
In our exercise, one of the key natural logarithm properties applied was \( e^{\ln a} = a \), which allowed the simplification of \( e^{\ln t^2} \) to \( t^2 \). This property reflects the fact that the exponential function and the natural logarithm are inverses.
Keep in mind the following properties when working with logarithms:

• \( \ln(ab) = \ln a + \ln b \)
• \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \)
• \( \ln(a^b) = b \ln a \)

These properties can simplify logarithmic expressions and make exponential manipulations more intuitive.
By understanding these natural logarithm properties, we can easily handle complex expressions in calculus, making it easier to solve equations like \( F(f(t), g(t)) = t^2 + e^t \).