Problem 6 - Applied Calculus

Question

For the functions \(f\) and \(g\), find (a) \(f(g(1))\) (b) \(g(f(1))\) (c) \(f(g(x))\) (d) \(g(f(x))\) (e) \(f(t) g(t)\) $$f(x)=1 / x, g(x)=3 x+4$$

Step-by-Step Solution

Answer

(a) \(\frac{1}{7}\), (b) 7, (c) \(\frac{1}{3x+4}\), (d) \(\frac{3}{x} + 4\), (e) \(\frac{3t+4}{t}\).

Step 1: Evaluate g(1)

First, substitute 1 into the function \(g(x) = 3x + 4\). This gives \(g(1) = 3(1) + 4 = 7\).

Step 2: Evaluate f(g(1))

Substitute the result from Step 1 into the function \(f(x) = \frac{1}{x}\). Since \(g(1) = 7\), then \(f(g(1)) = f(7) = \frac{1}{7}\).

Step 3: Evaluate f(1)

Substitute 1 into the function \(f(x) = \frac{1}{x}\). This gives \(f(1) = \frac{1}{1} = 1\).

Step 4: Evaluate g(f(1))

Substitute the result from Step 3 into the function \(g(x) = 3x + 4\). Since \(f(1) = 1\), then \(g(f(1)) = g(1) = 7\).

Step 5: Express f(g(x))

Substitute \(g(x) = 3x + 4\) into \(f(x) = \frac{1}{x}\). Thus, \(f(g(x)) = f(3x + 4) = \frac{1}{3x + 4}\).

Step 6: Express g(f(x))

Substitute \(f(x) = \frac{1}{x}\) into \(g(x) = 3x + 4\). Thus, \(g(f(x)) = g\left(\frac{1}{x}\right) = 3\left(\frac{1}{x}\right) + 4 = \frac{3}{x} + 4\).

Step 7: Calculate f(t) g(t)

Substitute \(f(t) = \frac{1}{t}\) and \(g(t) = 3t + 4\) into the expression \(f(t)g(t)\). Therefore, \(f(t)g(t) = \frac{1}{t} \times (3t + 4) = \frac{3t + 4}{t}\).

Key Concepts

Function Composition

Function composition is a key concept in algebra and calculus. It involves applying one function to the results of another function. Think of it as a chain reaction, where the output of one function becomes the input of the next. In this exercise, we're asked to find compositions like \(f(g(x))\) and \(g(f(x))\).

For example:

• To find \(f(g(x))\), you start by substituting the entire expression of \(g(x)\) into the function \(f\).
• If \(g(x) = 3x + 4\), then \(f(g(x))\) becomes \(f(3x + 4)\).
• Since \(f(x) = \frac{1}{x}\), substituting \(3x + 4\) gives \(f(3x + 4) = \frac{1}{3x + 4}\).

Remember, the order in which functions are composed matters. \(f(g(x))\) is not the same as \(g(f(x))\).
This is because you're performing different operations in a different order.
Be mindful of this while solving function composition problems.

Piecewise Functions

Piecewise functions are unique because they allow us to define different behaviors for different intervals of the input. Although they don't directly relate to this exercise, understanding piecewise functions can enhance your grasp of more complex function manipulation scenarios.

Imagine a function that calculates your parking fees based on time:

• For the first hour, it's $5.
• For any additional hour, it's $2 per hour.

This is a piecewise function because the rule for calculating the fee changes based on whether one hour or more has passed.
By breaking down situations into manageable pieces, these functions become versatile tools in mathematics, allowing us to tailor solutions to complex issues.

Calculus Problems

Solving calculus problems often involves understanding how composite functions and piecewise functions work within differentiation or integration contexts. In calculus, function compositions are used to apply the chain rule or derivatives of composed functions.

When we differentiate \(f(g(x))\), we're using the chain rule:

• Find the derivative of the outer function \(f\), leaving the inner function \(g(x)\) as is.
• Multiply it by the derivative of the inner function \(g(x)\).

Consider \(f = \frac{1}{u}\) and \(u = g(x)\), you differentiate \(f\) with respect to \(u\) and then multiply by the derivative of \(u\) with respect to \(x\).
This approach simplifies complex-looking calculus problems, allowing us to navigate seamlessly between composed and regular functions.