Problem 29
Question
For the following exercises, evaluate the function \(f\) at the indicated values \(f(-3), f(2), f(-a),-f(a), f(a+h)\). \(f(x)=\sqrt{2-x}+5\)
Step-by-Step Solution
Verified Answer
\(f(-3) = \sqrt{5} + 5\), \(f(2) = 5\), \(f(-a) = \sqrt{2+a} + 5\), \(-f(a) = -\sqrt{2-a} - 5\), \(f(a+h) = \sqrt{2-a-h} + 5\).
1Step 1: Evaluate f(-3)
To find \(f(-3)\), substitute \(-3\) for \(x\) in the function. The expression becomes:\[f(-3)=\sqrt{2-(-3)} + 5=\sqrt{2+3} + 5=\sqrt{5} + 5.\]
2Step 2: Evaluate f(2)
To find \(f(2)\), substitute \(2\) for \(x\) in the function. The expression becomes:\[f(2)=\sqrt{2-2} + 5=\sqrt{0} + 5=5.\]
3Step 3: Evaluate f(-a)
To find \(f(-a)\), substitute \(-a\) for \(x\) in the function. The expression becomes:\[f(-a)=\sqrt{2-(-a)} + 5=\sqrt{2+a} + 5.\]
4Step 4: Evaluate -f(a)
To find \(-f(a)\), first determine \(f(a)\), and then negate the result. The expression for \(f(a)\) is:\[f(a)=\sqrt{2-a} + 5,\] and negating this gives:\[-f(a) = -\left(\sqrt{2-a} + 5\right) = -\sqrt{2-a} - 5.\]
5Step 5: Evaluate f(a+h)
To find \(f(a+h)\), substitute \(a+h\) for \(x\) in the function. The expression becomes: \[f(a+h)=\sqrt{2-(a+h)} + 5 = \sqrt{2-a-h} + 5.\]
Key Concepts
Radical FunctionsAlgebraic ExpressionsFunction NotationSubstitution Method
Radical Functions
Radical functions are mathematical expressions that include a square root, cube root, or any higher root. The function used in our exercise, \( f(x) = \sqrt{2-x} + 5 \), is a radical function because it contains a square root term \( \sqrt{2-x} \). This term specifies that for any value of \( x \), you subtract \( x \) from 2, and then take the square root of the result.
Radicals can be tricky because they are not always defined for all real numbers. For square roots, the expression under the square root symbol (also known as the radicand) must be non-negative to produce a real number result. In our function, the radicand is \( 2-x \); thus, for real number solutions, \( x \) must be less than or equal to 2.
Radicals can be tricky because they are not always defined for all real numbers. For square roots, the expression under the square root symbol (also known as the radicand) must be non-negative to produce a real number result. In our function, the radicand is \( 2-x \); thus, for real number solutions, \( x \) must be less than or equal to 2.
- Radicals induce domain restrictions that are essential to identify before evaluating the function.
- Ensure the radicand is non-negative in the context of real numbers.
Algebraic Expressions
An algebraic expression consists of numbers, variables, and operations organized in a meaningful way. In our example, we have the algebraic expression \( \sqrt{2-x}+5 \). Here, "2-x" is the radicand inside the square root function, and "+5" is an operation added to the result of the square root.
Algebraic expressions are the building blocks of functions in algebra. They can be modified and simplified using rules of arithmetic and algebraic principles, such as combining like terms and the distributive property. In radical functions, simplifying expressions often involves resolving the radicand before taking the appropriate root. For example, evaluating \( f(-3) \), you modify the expression \( 2 - (-3) \) to \( 2 + 3 \), resulting in \( \sqrt{5} + 5 \).
Algebraic expressions are the building blocks of functions in algebra. They can be modified and simplified using rules of arithmetic and algebraic principles, such as combining like terms and the distributive property. In radical functions, simplifying expressions often involves resolving the radicand before taking the appropriate root. For example, evaluating \( f(-3) \), you modify the expression \( 2 - (-3) \) to \( 2 + 3 \), resulting in \( \sqrt{5} + 5 \).
- Understand the role of each component in an algebraic expression.
- Use arithmetic and algebraic laws to manipulate expressions.
Function Notation
Function notation is a way to denote functions using symbols. It provides a clear and systematic means to describe mathematical relationships between variables. In our example, \( f(x) = \sqrt{2-x} + 5 \) uses "\( f \)" to represent the function name, "\( x \)" to indicate the variable of the function, and the expression "\( \sqrt{2-x} + 5 \)" to describe the action taken on the input \( x \).
By using function notation, one can efficiently communicate the input-output relationship in mathematical terms. The notation \( f(a+h) \) implies substituting \( a+h \) in place of \( x \) within the function, illustrating how the function will evaluate a specific form of variable modification. By clearly understanding function notation, you can apply functions to different specific values and expressions effectively.
By using function notation, one can efficiently communicate the input-output relationship in mathematical terms. The notation \( f(a+h) \) implies substituting \( a+h \) in place of \( x \) within the function, illustrating how the function will evaluate a specific form of variable modification. By clearly understanding function notation, you can apply functions to different specific values and expressions effectively.
- Function notation simplifies the substitution and evaluation process.
- It encapsulates complex operations in a concise and understandable manner.
Substitution Method
The substitution method is a foundational technique for function evaluation where you replace variables with actual values or other expressions. In our example, to find \( f(-3) \), you substitute \(-3\) in place of \( x \) in the function \( f(x) = \sqrt{2-x} + 5 \). This process transforms the expression into \( \sqrt{5} + 5 \).
Substitution plays a crucial role as it allows you to evaluate functions at specific points and is a primary step in solving many algebraic problems. For instance, evaluating \( f(a+h) \) involves replacing \( x \) with \( a+h \), which results in \( \sqrt{2-a-h} + 5 \). This method provides the ability to handle changes in the variable efficiently and understand the function's behavior under different scenarios.
Substitution plays a crucial role as it allows you to evaluate functions at specific points and is a primary step in solving many algebraic problems. For instance, evaluating \( f(a+h) \) involves replacing \( x \) with \( a+h \), which results in \( \sqrt{2-a-h} + 5 \). This method provides the ability to handle changes in the variable efficiently and understand the function's behavior under different scenarios.
- It is essential for evaluating functions at given points.
- It simplifies complex algebraic manipulations.
Other exercises in this chapter
Problem 29
For the following exercises, find functions \(f(x)\) and \(g(x)\) so the given function can be expressed as \(h(x)=f(g(x))\). \(h(x)=\frac{4}{(x+2)^{2}}\)
View solution Problem 29
For the following exercises, find the average rate of change of each function on the interval specified. \(h(x)=5-2 x^{2}\) on [-2,4]
View solution Problem 30
For the following exercises, graph the given functions by hand. \(f(x)=-|x+4|-3\)
View solution Problem 30
For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions. \(m(t)=3+\sqrt{t+2}\)
View solution