Problem 5 - Thomas Calculus

Question

If \(f(x)=x+5\) and \(g(x)=x^{2}-3,\) find the following. $$(a) f(g(0)) \quad \text { b. } g(f(0)) $$ $$ (c)f(g(x)) \quad \text { d. } g(f(x)) $$ $$ \begin{array}{ll}{\text { e. } f(f(-5))} & {\text { f. } g(g(2))} \\ {\text { g. } f(f(x))} & {\text { h. } g(g(x))}\end{array} $$

Step-by-Step Solution

Answer

(a) 2, (b) 22, (c) x^2 + 2, (d) x^2 + 10x + 22, (e) 5, (f) -2, (g) x + 10, (h) x^4 - 6x^2 + 6

Step 1: Solve for f(g(0))

First, find \( g(0) \) by substituting 0 into \( g(x) = x^2 - 3 \). This yields \( g(0) = 0^2 - 3 = -3 \). Next, substitute \( -3 \) into \( f(x) = x + 5 \). Therefore, \( f(g(0)) = f(-3) = -3 + 5 = 2 \).

Step 2: Solve for g(f(0))

First, find \( f(0) \) by substituting 0 into \( f(x) = x + 5 \). This yields \( f(0) = 0 + 5 = 5 \). Next, substitute \( 5 \) into \( g(x) = x^2 - 3 \). Therefore, \( g(f(0)) = g(5) = 5^2 - 3 = 25 - 3 = 22 \).

Step 3: Solve for f(g(x))

Substitute \( g(x) = x^2 - 3 \) into \( f(x) = x + 5 \) to get \( f(g(x)) = (x^2 - 3) + 5 = x^2 + 2 \).

Step 4: Solve for g(f(x))

Substitute \( f(x) = x + 5 \) into \( g(x) = x^2 - 3 \) to get \( g(f(x)) = (x + 5)^2 - 3 \). Expand it to get \( g(f(x)) = (x^2 + 10x + 25) - 3 = x^2 + 10x + 22 \).

Step 5: Solve for f(f(-5))

First, find \( f(-5) \) by substituting \(-5\) into \( f(x) = x + 5 \). This yields \( f(-5) = -5 + 5 = 0 \). Then substitute \( 0 \) back into \( f(x) \) to get \( f(f(-5)) = f(0) = 5 \).

Step 6: Solve for g(g(2))

First, find \( g(2) \) by substituting \(2\) into \( g(x) = x^2 - 3 \). This yields \( g(2) = 4 - 3 = 1 \). Then substitute \( 1 \) back into \( g(x) \) to get \( g(g(2)) = g(1) = 1^2 - 3 = -2 \).

Step 7: Solve for f(f(x))

Substitute \( f(x) = x + 5 \) into itself to get \( f(f(x)) = f(x + 5) = (x + 5) + 5 = x + 10 \).

Step 8: Solve for g(g(x))

Substitute \( g(x) = x^2 - 3 \) into itself to get \( g(g(x)) = g(x^2 - 3) = (x^2 - 3)^2 - 3 \). Simplify it to \( x^4 - 6x^2 + 9 - 3 = x^4 - 6x^2 + 6 \).

Key Concepts

Function Composition

Function composition involves the process of applying one function to the results of another. This idea is fundamental in various areas of mathematics, especially when dealing with complex expressions and transformations. Consider functions as machines: you feed a value into one machine, and its output is used as the input for another machine.

For example, if you have two functions, \( f(x) \) and \( g(x) \), composing \( f \) with \( g \), denoted as \( f(g(x)) \), means plugging the output of \( g(x) \) into \( f(x) \). This way, the composition creates a single function from two separate functions.

Function composition can be straightforward when dealing with numerical values, formulas, or expressions. Solving these requires substituting one function into the other and simplifying the resultant expression.

Polynomial Functions

Polynomial functions are algebraic expressions consisting of variables raised to powers, combined using addition, subtraction, and multiplication. They play a crucial role in mathematics, representing a wide variety of natural phenomena.

These functions can have varying degrees, determined by the highest power of the variable within them. For instance, \( g(x) = x^2 - 3 \) is a polynomial of degree 2. It’s called a quadratic function because its highest power is 2.

Understanding polynomial functions helps with evaluating function compositions and performing operations such as addition, subtraction, differentiation, and integration on them. Calculations involving these functions often lead to further insights into how complex systems behave.

• Degree: The highest exponent of the variable.
• Terms: The separated components in the expression (e.g., \(x^2, -3\)).
• Coefficients: Numbers multiplying the variable's powers.

Step-by-step Solutions

Following a step-by-step approach is beneficial for solving mathematical problems, especially when it comes to function composition. It helps in systematically breaking down complex problems into manageable parts.

Let’s consider an example: to solve \( f(g(0)) \), first find \( g(0) \) which simplifies simplifying \( g(0) = 0^2 - 3 = -3 \). This result is then plugged into \( f \), giving \( f(-3) = -3 + 5 = 2 \). This breakdown ensures clarity and prevents errors.

Approaching problems step-by-step also helps in identifying patterns and making connections between different mathematical concepts. Each step provides an opportunity to verify calculations and adjust any mistakes early in the process, thereby reinforcing fundamental skills such as algebraic manipulation and polynomial function comprehension.

• Identify the innermost function first.
• Substitute the result into the outer function.
• Simplify and verify results.

Algebraic Manipulations

Algebraic manipulation is the method of rearranging and simplifying expressions to solve problems or reveal other forms of the expressions. It’s an essential skill for handling compositions of functions and working with polynomial expressions.

This process includes expanding brackets, combining like terms, factorizing expressions, and performing arithmetic operations. When working with compositions such as \( g(f(x)) = (x + 5)^2 - 3 \), expanding \( (x + 5)^2 \) to \( x^2 + 10x + 25 \) is a step that employs these basic manipulation techniques. The result is further simplified by combining terms and performing subtraction.

Algebraic manipulation ensures flexibility and efficiency in handling problems and can simplify seemingly complex tasks into understandable sections. It is also vital in verifying results from composed functions, determining equivalences, and transforming expressions in various ways beneficial for solving different mathematics problems.

• Expanding: Distributing terms across brackets.
• Combining: Simplifying by merging like terms.
• Factorizing: Rewriting expressions as products of factors.