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Chapter 8: Problem 59

Find \(h(5)\) and \(h(-2) .\) See Example 4. $$ h(x)=|x|+2 $$

Short Answer

Expert verified
\( h(5) = 7 \) and \( h(-2) = 4 \).

Step by step solution

01

Understand the Function

The function given is \( h(x) = |x| + 2 \). This means you take the absolute value of \( x \) and then add 2 to the result.
02

Evaluate \( h(5) \)

Substitute \( x = 5 \) into the function. Calculate the absolute value of 5, which is 5, and then add 2. So, \( h(5) = |5| + 2 = 5 + 2 = 7 \).
03

Evaluate \( h(-2) \)

Substitute \( x = -2 \) into the function. Calculate the absolute value of -2, which is 2, and then add 2. So, \( h(-2) = |-2| + 2 = 2 + 2 = 4 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is a fundamental concept in mathematics. It involves substituting a given value into a function. This helps us find the corresponding output of that function. For example, if you have a function like \( h(x) = |x|+2 \), evaluating \( h(5) \) means substituting 5 into \( x \):
  • Start by replacing \( x \) with 5: \( h(5) = |5|+2 \).
  • Find the absolute value: \( |5| = 5 \).
  • Add 2 to the result: \( 5+2 = 7 \). Thus, \( h(5) = 7 \).
It's a simple process where you're finding the output by performing operations specified by the function's rule. Evaluating functions like this is crucial in both understanding algebra and in solving real-world problems.
Absolute Value
The concept of absolute value is vital in algebra and mathematics in general. The absolute value of a number refers to its distance from 0 on the number line, without considering direction. Let's break it down:
  • The absolute value is always non-negative.
  • For any positive number \( x \), \( |x| = x \).
  • For any negative number \( x \), \( |x| = -x \) (which makes it positive).
In our function \( h(x) = |x| + 2 \), when calculating \( h(-2) \), you take the absolute value of \(-2\), which is 2. This step eliminates any negative sign, making calculations straightforward. This attribute makes absolute value particularly useful for many real-world applications, such as measuring distances and error handling in data.
Algebraic Expressions
Algebraic expressions are combinations of variables, numbers, and operations such as addition, subtraction, multiplication, and division. They form the basis of algebra and allow us to describe patterns and relationships between numbers. An example of an algebraic expression is \( h(x) = |x| + 2 \).Here's how it works:
  • \( h(x) \) refers to the output based on input \( x \).
  • The term \( |x| \) finds the absolute value of \( x \).
  • Adding 2 is the last operation to get the final output.
Algebraic expressions with absolute values are particularly useful because they always give a positive solution. This ensures consistent results when problem-solving, making them invaluable in both academic contexts and practical applications.

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Most popular questions from this chapter

Perform the operations and simplify, if possible. See Example 5 $$\frac{2 p^{2}-5 p-3}{p^{2}-9} \cdot \frac{2 p^{2}+5 p-3}{2 p^{2}+5 p+2}$$

Solve each equation. $$-10|16 h+4|-3=-3$$

Perform the operations and simplify, if possible. See Example 9. $$\frac{x+2}{x+5}-\frac{x-3}{x+7}$$

Find the domain of \(f(x)=\frac{1}{5[9(x-2)-6(x-3)+3]}\)

Perform the operations and simplify, if possible. See Example 7. $$\frac{3 y-2}{2 y+6}-\frac{2 y-5}{2 y+6}$$
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