Problem 3 - 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) \(f(t) g(t)\) $$f(x)=x^{2}, g(x)=x+1$$

Step-by-Step Solution

Answer

(a) 4, (b) 2, (c) \(x^2 + 2x + 1\), (d) \(x^2 + 1\), (e) \(t^3 + t^2\)

Step 1: Evaluate g(1)

To find \(f(g(1))\), we need to first determine \(g(1)\). By using the function \(g(x) = x + 1\), substitute \(x = 1\) to get \(g(1) = 1 + 1 = 2\).

Step 2: Evaluate f(g(1))

Now, with \(g(1) = 2\) found in Step 1, substitute this into function \(f\). Since \(f(x) = x^2\), we have \(f(g(1)) = f(2) = 2^2 = 4\).

Step 3: Evaluate f(1)

To find \(g(f(1))\), we need to first determine \(f(1)\). Use \(f(x) = x^2\) and substitute \(x = 1\) to get \(f(1) = 1^2 = 1\).

Step 4: Evaluate g(f(1))

With \(f(1) = 1\) as found in Step 3, substitute into \(g(x)\). Using \(g(x) = x + 1\), we have \(g(f(1)) = g(1) = 1 + 1 = 2\).

Step 5: Evaluate f(g(x))

For \(f(g(x))\), substitute \(g(x) = x + 1\) into the function \(f\). This becomes \(f(g(x)) = f(x+1) = (x+1)^2\).

Step 6: Simplify f(g(x))

Simplify \((x+1)^2\) using the binomial expansion: \((x+1)^2 = x^2 + 2x + 1\). Thus, \(f(g(x)) = x^2 + 2x + 1\).

Step 7: Evaluate g(f(x))

For \(g(f(x))\), substitute \(f(x) = x^2\) into the function \(g\). This gives \(g(f(x)) = g(x^2) = x^2 + 1\).

Step 8: Evaluate f(t) * g(t)

To find \(f(t) g(t)\), first use the separate functions: \(f(t) = t^2\) and \(g(t) = t + 1\). Multiply them to get \(f(t) g(t) = t^2 (t + 1)\).

Step 9: Simplify f(t) * g(t)

Distribute the multiplication: \(t^2 imes (t + 1) = t^3 + t^2\). Thus, \(f(t) g(t) = t^3 + t^2\).

Key Concepts

Composite Functions

One of the fundamental aspects of understanding functions is learning how to work with composite functions. A composite function is created when one function is applied to the result of another function. The notation for composite functions is usually written as \( f(g(x)) \) or \( g(f(x)) \), where the function \( g(x) \) is applied to \( x \) first, and then \( f \) is applied to that output.
Imagine you have two functions: \( f(x) = x^2 \) and \( g(x) = x + 1 \). Composing these means:

• Evaluating \( g(x) \) first, and using its result in \( f \).
• Or, Evaluating \( f(x) \) first, then using its result in \( g \).

For example, to find \( f(g(x)) \), substitute \( g(x) = x + 1 \) into \( f \), resulting in \( f(x+1) = (x+1)^2 \). Similarly, \( g(f(x)) \) evaluates to \( g(x^2) = x^2 + 1 \). Understanding composite functions helps in analyzing complex mathematical models where multiple functional relationships are involved.

Function Evaluation

Function evaluation involves plugging in specific values for the function's variable and determining the output. The process may differ slightly depending on the nature of the function, but the core idea remains the same: find a solution that follows the functional rule.
Let's illustrate this with the functions \( f(x) = x^2 \) and \( g(x) = x + 1 \).To evaluate \( f(g(1)) \), first calculate \( g(1) \). Substitute 1 into \( g \), giving \( g(1) = 2 \).
Then use this result in \( f \), so \( f(g(1)) = f(2) = 4 \).Similarly, to find \( g(f(1)) \), evaluate \( f(1) = 1 \).
Insert this into \( g \) to find \( g(f(1)) = 2 \).
Function evaluation is a crucial step in determining outputs for specific inputs, helping us understand how functions behave.

Polynomial Functions

Polynomial functions are expressions involving variables raised to whole number exponents with constant coefficients. They can take many forms, including linear, quadratic, cubic, and more complicated multivariable functions.
In the example provided, both \( f(x) = x^2 \) and \( g(x) = x + 1 \) are polynomial functions:

• \( f \) is a quadratic polynomial, as it has the form \( ax^2 + bx + c \).
• \( g \) is a linear polynomial of the form \( ax + b \).

Polynomial functions are foundational in algebra due to their versatile nature in modeling real-world phenomena. Understanding how various polynomials interact through operations like addition, subtraction, multiplication, or composition is an essential skill, enabling the solvers to simplify and solve multi-step problems.

Arithmetic Operations on Functions

Arithmetic operations on functions involve combining functions through addition, subtraction, multiplication, and division. In some cases, you can use these operations to form new functions or simplify expressions.
For instance, if you have \( f(t) = t^2 \) and \( g(t) = t + 1 \), multiplying \( f(t) \) and \( g(t) \) involves:

• First, evaluating each function separately, \( f(t) = t^2 \) and \( g(t) = t + 1 \).
• Then, applying the multiplication: \( f(t) g(t) = t^2(t + 1) \).
• Simplifying the result to get \( t^3 + t^2 \).

This technique allows for the expansion of expressions, playing an important role in calculus and algebra. By mastering arithmetic operations on functions, you can handle more complex operations and solve a wide variety of mathematical problems effectively.