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Chapter 1: Problem 26

Evaluating a Function In Exercises \(21-32\) , evaluate (if possible) the function at each specified value of the independent variable and simplify. $$ \begin{array}{l}{f(x)=\sqrt{x+8}+2} \\ {\text { (a) } f(-8) \quad \text { (b) } f(1)}\end{array} \quad \text { (c) } f(x-8) $$

Short Answer

Expert verified
For the given exercise, the evaluated function for x being -8, 1 and \(x-8\) are 2, 5, and \(\sqrt{x}+2\) respectively.

Step by step solution

01

Substitute x with -8

Without considering about the existence of the mathematical concept, you simply plug in -8 to the function: \(f(-8) = \sqrt{-8 + 8} + 2 = \sqrt{0} + 2 = 0 + 2 = 2\)
02

Substitute x with 1

When substituting x with 1 in \(f(x) = \sqrt{x+8} + 2\), the function becomes: \(f(1) = \sqrt{1+8}+2 = \sqrt{9} + 2 = 3 + 2 = 5\)
03

Substitute x with (x-8)

Finally, replacing x with (x-8) in \(f(x) = \sqrt{x+8} + 2\), yields: \(f(x-8) = \sqrt{x-8+8}+2 = \sqrt{x}+2\)

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Key Concepts

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

Understanding the Independent Variable
Functions are mathematical expressions that show the relationship between two quantities. One quantity is the dependent variable, and the other is the independent variable. In the expression \(f(x) = \sqrt{x+8} + 2\), \(x\) is the independent variable.
The independent variable is the input value of the function. It is "independent" because you can choose any value for it, giving you the freedom to see how the function behaves as \(x\) changes.
  • Choose values for \(x\) to explore how they affect the function's output.
  • Examples of independent values are numbers like \(-8, 1,\) or even algebraic expressions like \(x-8\).

In summary, understanding and using the independent variable effectively allows us to evaluate functions and see how different inputs change the results.
Simplifying Functions
Function simplification is the process of breaking down an expression into a simpler form while keeping its value unchanged. This makes it easier to calculate results without changing the function's fundamental nature.
When the function \(f(x) = \sqrt{x+8} + 2\) is evaluated with specific values, simplification helps in verifying the outcomes efficiently. To simplify is to follow through with the operations gradually:
  • Consider \(f(-8)\): Calculate each component, \(\sqrt{-8 + 8} = \sqrt{0}\).
  • Continue to simplify: Adding \(2\) gives the final value.

The goal here is to ensure each step of the calculation remains clear and correct. By simplifying, you easily verify answers and ensure accuracy.
Implementing the Substitution Method
The substitution method involves replacing the independent variable with specific values or expressions. In essence, it plugs these substitutive values directly into the function.
Using \(f(x) = \sqrt{x+8} + 2\), the substitution method enables you to evaluate the function at given points by:
  • Substituting \(-8\) for \(x\): \(f(-8)\) becomes \(\sqrt{-8+8} + 2\) which simplifies to \(2\).
  • Substituting \(1\) for \(x\): Similarly, \(f(1) = \sqrt{1+8} + 2\) results in \(5\).
  • Substituting \(x-8\) for \(x\): Leads to a slightly different kind of substitution \(f(x-8) = \sqrt{x}+2\), simplifying to a new expression.

This method is powerful for solving various scenarios in mathematical expressions by efficiently evaluating different inputs in the function.

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Most popular questions from this chapter

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Composition with lnverses In Exercises \(89-92\) , use the functions \(f(x)=x+4\) and \(g(x)=2 x-5\) to find the specified function. $$(g \circ f)^{-1}$$

Intercept Form of the Equation of a line, use the intercept form to find the equation of the line with the given intercepts. The intercept form of the equation of a line with intercepts \((a, 0)\) and \((0, b)\) is $$\frac{x}{a}+\frac{y}{b}=1, a \neq 0, b \neq 0$$ $$\begin{array}{l}{x \text { -intercept: }\left(-\frac{1}{6}, 0\right)} \\ {y \text { -intercept: }\left(0,-\frac{2}{3}\right)}\end{array}$$

Parallel and Perpendicular Lines, determine whether the lines are parallel, perpendicular, or neither. $$\begin{array}{l}{L_{1} : y=\frac{1}{2} x-3} \\ {L_{2} : y=-\frac{1}{2} x+1}\end{array}$$
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