Problem 87
Question
For the following exercises, find the composition when \(f(x)=x^{2}+2\) for all \(x \geq 0\) and \(g(x)=\sqrt{x-2}\). $$(f \circ g)(6) ;(g \circ f)(6)$$
Step-by-Step Solution
Verified Answer
Both \((f \circ g)(6)\) and \((g \circ f)(6)\) equal 6.
1Step 1: Understanding the Composition
To find the compositions \((f \circ g)(x)\) and \((g \circ f)(x)\), we substitute one function into the other. For \((f \circ g)(x)\), we replace \(x\) in \(f(x)\) with \(g(x)\). For \((g \circ f)(x)\), we replace \(x\) in \(g(x)\) with \(f(x)\).
2Step 2: Finding \((f \circ g)(6)\)
First, calculate \(g(6)\). Since \(g(x) = \sqrt{x - 2}\), we have: \[ g(6) = \sqrt{6 - 2} = \sqrt{4} = 2. \] Next, substitute this result into \(f(x)\): \[ f(2) = 2^2 + 2 = 4 + 2 = 6. \] So, \((f \circ g)(6) = 6\).
3Step 3: Finding \((g \circ f)(6)\)
First, calculate \(f(6)\). Since \(f(x) = x^2 + 2\), we have: \[ f(6) = 6^2 + 2 = 36 + 2 = 38. \] Next, substitute this result into \(g(x)\): \[ g(38) = \sqrt{38 - 2} = \sqrt{36} = 6. \] So, \((g \circ f)(6) = 6\).
Key Concepts
Inverse FunctionsDomain and RangeFunction Evaluation
Inverse Functions
Inverse functions can be thought of as undoing what a function does. If you have a function that takes you from point A to point B, the inverse function takes you back from point B to point A. In the context of function pairs, if you apply a function and its inverse one after the other, you'll end up where you started. This property is sometimes referred to as the 'undo' property of inverses.
However, not all functions have inverses. A function must be one-to-one (bijective) to have an inverse. One easy way to remember this is that each output of the function must be unique and only have one corresponding input. This ensures that when the function is reversed (i.e., the inverse is applied), each input is also unique. When studying inverse functions, it's common to find the inverse by swapping the roles of the function’s inputs and outputs, and then solving for the new input variable using algebraic methods.
However, not all functions have inverses. A function must be one-to-one (bijective) to have an inverse. One easy way to remember this is that each output of the function must be unique and only have one corresponding input. This ensures that when the function is reversed (i.e., the inverse is applied), each input is also unique. When studying inverse functions, it's common to find the inverse by swapping the roles of the function’s inputs and outputs, and then solving for the new input variable using algebraic methods.
Domain and Range
When discussing functions, the concepts of domain and range are crucial. The domain of a function is the set of all possible inputs (x-values) that the function can accept. Meanwhile, the range of a function is the set of all possible outputs (y-values) that can be generated by the function.
For example, in the function given, \(f(x) = x^2 + 2\), the domain is \(x \geq 0\) because the function is defined only for non-negative values of \(x\). The range of \(f(x)\) starts from 2 and can go to infinity because squaring any non-negative number and adding two will always be greater than or equal to 2.
The function \(g(x) = \sqrt{x - 2}\) has a domain that starts from \(x = 2\) onwards because we cannot take the square root of a negative number in the real number system. The range of this function is \(y \geq 0\) because the square root of any non-negative number is also non-negative.
Understanding domain and range is crucial, especially when composing functions, as one function's range must fall within the next function's domain, otherwise, the composition is not possible.
For example, in the function given, \(f(x) = x^2 + 2\), the domain is \(x \geq 0\) because the function is defined only for non-negative values of \(x\). The range of \(f(x)\) starts from 2 and can go to infinity because squaring any non-negative number and adding two will always be greater than or equal to 2.
The function \(g(x) = \sqrt{x - 2}\) has a domain that starts from \(x = 2\) onwards because we cannot take the square root of a negative number in the real number system. The range of this function is \(y \geq 0\) because the square root of any non-negative number is also non-negative.
Understanding domain and range is crucial, especially when composing functions, as one function's range must fall within the next function's domain, otherwise, the composition is not possible.
Function Evaluation
When we evaluate a function, we're essentially finding the output value that corresponds to a particular input value. This is done by substituting the input value into the function's equation.
Let's consider the given function \(f(x) = x^2 + 2\). If we want to evaluate \(f\) at \(x = 6\), we substitute 6 in place of \(x\), which gives us \(6^2 + 2 = 38\). This tells us that when the input is 6, the output is 38.
Similarly, for the function \(g(x) = \sqrt{x - 2}\), evaluating it at some \(x\) means subtracting 2 from \(x\) and then taking the square root of the result. For example, if \(x = 6\), then \(g(6) = \sqrt{6 - 2} = \sqrt{4} = 2\).
Function evaluation is a fundamental skill in algebra and calculus that helps us understand how functions behave and how to use them to model real-world situations efficiently. It enables us to find specific values quickly or to determine the function's behavior over a range of inputs.
Let's consider the given function \(f(x) = x^2 + 2\). If we want to evaluate \(f\) at \(x = 6\), we substitute 6 in place of \(x\), which gives us \(6^2 + 2 = 38\). This tells us that when the input is 6, the output is 38.
Similarly, for the function \(g(x) = \sqrt{x - 2}\), evaluating it at some \(x\) means subtracting 2 from \(x\) and then taking the square root of the result. For example, if \(x = 6\), then \(g(6) = \sqrt{6 - 2} = \sqrt{4} = 2\).
Function evaluation is a fundamental skill in algebra and calculus that helps us understand how functions behave and how to use them to model real-world situations efficiently. It enables us to find specific values quickly or to determine the function's behavior over a range of inputs.
Other exercises in this chapter
Problem 86
For the following exercises, graph \(y=\sqrt[3]{x}\) on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. $$
View solution Problem 87
Find the composition when \(f(x)=x^{2}+2\) for all \(x \geq 0\) and \(g(x)=\sqrt{x-2}\). $$ (f \circ g)(6) ;(g \circ f)(6) $$
View solution Problem 88
Find the composition when \(f(x)=x^{2}+2\) for all \(x \geq 0\) and \(g(x)=\sqrt{x-2}\). $$ (g \circ f)(a) ;(f \circ g)(a) $$
View solution Problem 88
For the following exercises, find the composition when \(f(x)=x^{2}+2\) for all \(x \geq 0\) and \(g(x)=\sqrt{x-2}\). $$(g \circ f)(a) ;(f \circ g)(a)$$
View solution