Problem 48 - Algebra 2

Question

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

Step-by-Step Solution

Answer

0

Step 1: Evaluate h(-1)

To find the value of \(h(-1)\), substitute \(-1\) into the function \(h(x) = x^2\): \[ h(-1) = (-1)^2 = 1 \] So, \( h(-1) = 1 \).

Step 2: Substitute into g(x)

Now we need to find \(g[h(-1)]\), which means substituting the result from Step 1 into \(g(x)\). Since \(h(-1) = 1\), we need to evaluate \(g(1)\): \[ g(x) = x - 1 \] Substitute \(x = 1\): \[ g(1) = 1 - 1 = 0 \]

Key Concepts

Function Evaluation

Function evaluation is the process of finding the output of a function for a particular input value. It involves substituting the input value into the function's formula and performing the operations indicated. Given a function, denoted as typically \(f(x)\), evaluating the function at a specific point \(x = a\) involves replacing \(x\) with \(a\) in the equation. For example, if we have the function \(h(x) = x^2\), and we are tasked with finding \(h(-1)\), we replace \(x\) with \(-1\) to find the result:

• Use \(h(x) = x^2\) as the function.
• Enter \(-1\) as the input value: \(h(-1) = (-1)^2 = 1\).

This computation shows how the value of \(h(-1)\) is determined to be \(1\). Evaluating functions in this manner is a consistent method used across various types of functions, from polynomials to more complex equations.

Substitution Method

The substitution method is a powerful technique used in mathematics that involves replacing a variable or expression with another value to simplify the problem-solving process. It's especially helpful when dealing with composite functions, where you apply one function to the result of another function. Let's break it down using our original exercise.

• First, evaluate the inner function, as shown with \(h(x)\):\[ h(-1) = 1 \]
• Once the result is known, substitute it into the next function, here with \(g(x)\):\[ g(1) = 1 - 1 = 0 \]

Here, the result from \(h(-1)\) is substituted directly into \(g(x)\), making it easy to evaluate \(g[h(-1)]\). This method streamlines solving complex function compositions by breaking them into simpler, sequential steps.

Polynomial Functions

Polynomial functions are mathematical expressions that consist of variables raised to whole number powers with constant coefficients. These functions are fundamental in algebra due to their straightforward structure and flexibility. For instance, \(h(x) = x^2\) is a polynomial function where the highest power of \(x\) is 2, making it a quadratic function. Polynomial functions can take forms like:

• Constant: \(f(x) = c\)
• Linear: \(f(x) = ax + b\)
• Quadratic: \(f(x) = ax^2 + bx + c\)

In the original exercise, evaluating the polynomial \(h(x) = x^2\) is straightforward: substitute the input into the expression and compute. For \(x = -1\), this results in \(h(-1) = 1\). Polynomial functions lay the groundwork for understanding more advanced algebraic concepts and still apply to various real-world problems.