Problem 89 - Algebra 2

Question

Let \(g(x)=x-3\) and \(h(x)=x^{2}+6 .\) Find \((h \circ g)(1)\) $$\begin{array}{llll}{\text { A. }-14} & {\text { B. } 4} & {\text { C. } 5} & {\text { D. } 10}\end{array}$$

Step-by-Step Solution

Answer

The short answer is D. 10.

Step 1: Evaluate function \(g\)

First, substitute \(1\) into function \(g(x)\) which is \(x-3\) obtaining \(g(1) = 1-3 = -2\).

Step 2: Substitute result into function \(h\)

Next, substitute the result from step 1, \(-2\), into function \(h(x) = x^{2}+6\), we get \(h(-2) = (-2)^{2}+6 = 4+6 = 10\).

Step 3: Interpret the result

The result from step 2 is the answer to the problem, therefore \((h \circ g)(1) = 10\) . The correct answer is D. 10.

Key Concepts

Understanding Function Composition

Function composition involves applying one function to the results of another function. It is denoted by \((h \circ g)(x)\), which is read as "\(h\) composed with \(g\)" of \(x\). In simpler terms:

• Evaluate the inner function first. This means substituting the given value into the innermost function.
• Take the result and substitute it into the next function.

For the problem \((h \circ g)(1)\), it means we first evaluate \(g(1)\) and then use that result in \(h\). This process streamlines calculations by leveraging the output of one function as the input of another.

Evaluating Functions Step-by-Step

Evaluating a function means finding the value of that function for a specific input. Here is how we evaluate based on our example:

• First, take the input value associated with the variable, which in the problem is \(x = 1\).
• Plug this input into the function \(g(x) = x - 3\). When \(x = 1\), \(g(1) = 1 - 3 = -2\).

Next, we evaluate function \(h(x) = x^2 + 6\) using the result from \(g(x)\). Substituting \(-2\) into \(h(x)\) gives \(h(-2) = (-2)^2 + 6 = 4 + 6 = 10\). Thus, evaluating simply involves calculating the output for a given input by substituting values.

Exploring Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operation symbols. They represent real-world quantities and can be manipulated to solve problems.

• An algebraic expression could be as simple as \(x - 3\) or \(x^2 + 6\), seen in functions \(g(x)\) and \(h(x)\) respectively.
• Each expression consists of terms: units of the expression separated by addition or subtraction, such as \(x\) and \(-3\) in \(x - 3\).
• Expressions can be simplified by performing operations like addition, subtraction, multiplication, and division based on standard algebraic rules.

To solve expressions within functions, substitute the variable with a specific value. Performing the correct operations then reveals the value of the whole expression. For instance, substituting and calculating produce answers required for further function evaluations like in the problem scenario.