Problem 37 - Algebra 2

Question

If \(f(x)=4 x, g(x)=2 x-1,\) and \(h(x)=x^{2}+1,\) find each value. $$ f[h(-4)] $$

Step-by-Step Solution

Answer

The value of \( f[h(-4)] \) is 68.

Step 1: Evaluate h(-4)

Let's begin by finding the value of \( h(-4) \). The function \( h(x) \) is defined as \( h(x) = x^2 + 1 \). So, \( h(-4) = (-4)^2 + 1 = 16 + 1 = 17 \).

Step 2: Evaluate f[h(-4)]

Now that we know \( h(-4) = 17 \), we need to find \( f(17) \). The function \( f(x) \) is defined as \( f(x) = 4x \). Substituting \( 17 \) into \( f(x) \), we get \( f(17) = 4 \times 17 = 68 \).

Key Concepts

Polynomial Functions

Polynomial functions are a major cornerstone in algebra, often appearing in various forms such as linear, quadratic, and more complex polynomials. In this exercise, we are introduced to two polynomial functions, specifically linear polynomials. Linear polynomial functions have the general form of \( ax + b \), where \( a \) and \( b \) are constants. For instance, both \( f(x) = 4x \) and \( g(x) = 2x - 1 \) are linear because their highest power of \( x \) is 1.

However, not all functions discussed here are linear. The function \( h(x) = x^2 + 1 \) is a quadratic polynomial because it has the highest power of 2, signifying a curve known as a parabola when graphed. Understanding these differences in degree is crucial:

• Linear functions: degree 1
• Quadratic functions: degree 2
• Higher degrees lead to more complex polynomials

Polynomials are used to model a variety of real-world phenomena, making them incredibly powerful in both theoretical and practical applications.

Function Evaluation

Function evaluation involves substituting a specific value of \( x \) into a function and computing the result. This is a fundamental skill in algebra as it applies across different types of functions, including polynomials. To solve the given exercise, we first evaluate \( h(-4) \) using the function \( h(x) = x^2 + 1 \). This means replacing \( x \) with \(-4\) and calculating:

• \( h(-4) = (-4)^2 + 1 = 16 + 1 = 17 \)

Next, we use this result to find \( f[ h(-4) ] \), which involves substituting \( 17 \) into \( f(x) = 4x \):

• \( f(17) = 4 \times 17 = 68 \)

This two-step process exemplifies function composition, where the output of one function becomes the input to another. It’s crucial to handle each step methodically for accuracy.

Algebraic Expressions

Algebraic expressions consist of variables, numbers, and operations. They form the building blocks of functions. In our exercise, the expressions \( f(x) = 4x \) and \( h(x) = x^2 + 1 \) contain distinct algebraic elements and operations - multiplication for \( f(x) \) and a combination of squaring, addition for \( h(x) \).

When working with algebraic expressions:

• Identify all parentheses and apply the order of operations correctly
• Understand each element (coefficients, variables, constants) and how they interact
• Keep expressions simplified and organized to avoid errors

Combining these expressions through composition, like in the exercise where \( f(h(x)) \) was assessed, requires careful attention to the resulting algebraic expression after inserting one function into another.