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Chapter 1: Problem 73

Find; a. \((f \circ g)(x)\) b. the domain of \(f \circ g\) $$f(x)=x^{2}+4, g(x)=\sqrt{1-x}$$

Short Answer

Expert verified
a. The composite function \(f \circ g(x) = 5 - x\).\nb. The domain of \(f \circ g(x)\) is all \(x\) such that \(x \leq 1\).

Step by step solution

01

Find the Composite Function

Let's first find the composite function \(f \circ g(x)\). Since \(f(x) = x^{2} + 4\) and \(g(x) = \sqrt{1 - x}\), we have \(f \circ g(x) = f(g(x)) = (\sqrt{1 - x})^{2} + 4 = 1 - x + 4 = 5 - x.\)
02

Find the Domain of \(f \circ g\)

To find the domain of the composite function \(f \circ g\), consider the domain of \(g\) first. The function \(g(x) = \sqrt{1 - x}\) is defined for all \(x\) such that \(1 - x \geq 0\), which implies \(x \leq 1\). As the function \(f(x) = x^{2} + 4\) is defined for all real numbers, all values that function \(g(x)\) can take are in the domain of \(f\). Hence, the domain of \(f \circ g\) is all \(x\) such that \(x \leq 1\).

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Key Concepts

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

Domain of a Function
In mathematics, the domain of a function is the set of all possible input values for which the function is defined. It's crucial to determine the domain because it tells us the boundaries within which a function can operate effectively. To find the domain of a composite function such as \(f \circ g\), we need to first look at the domain of the inner function 鈥� in this case, \(g(x)\).

  • For the function \(g(x) = \sqrt{1 - x}\), the expression under the square root, \(1 - x\), must be non-negative for the square root to be valid in the real number system.
  • This means \(1 - x \geq 0\) or \(x \leq 1\). Therefore, the domain of \(g(x)\) is \((-\infty, 1]\).
The outer function \(f(x) = x^2 + 4\) is defined for all real numbers, which broadens the overall possibility initially. However, since \(g(x)\) must be defined first to input into \(f(x)\), the domain of \(f \circ g\) remains \(x \leq 1\).
Function Composition
Function composition is an operation that takes two functions, \(f(x)\) and \(g(x)\), and creates a new function \(f \circ g(x)\) by applying \(g\) and then \(f\). It's like chaining separate actions:

  • First substitute \(x\) into \(g(x)\).
  • Then take the result of \(g(x)\) and substitute it into \(f(x)\).
For the given functions, \(f(x) = x^2 + 4\) and \(g(x) = \sqrt{1-x}\), the composite \(f \circ g(x)\) involves the steps:1. Calculate \(g(x) = \sqrt{1-x}\).2. Substitute \(g(x)\) into \(f(x)\), resulting in \(f(g(x)) = (\sqrt{1-x})^2 + 4\).3. Simplify the expression to get \(5 - x\).

This shows us that the operations within function composition need to be handled sequentially.
Radical Functions
Radical functions are functions that include roots, such as square roots or cube roots. Understanding how these functions work is crucial because their domains are often restricted.

  • For square root functions like \(\sqrt{x}\), the expression under the root must not be negative, since the square root of a negative number isn't real (unless you're working with complex numbers).
  • This restriction in the expression \(g(x) = \sqrt{1-x}\) means \(1-x\) must be greater than or equal to zero, so \(x\) must be less than or equal to 1.
Thus, radical functions play a key role in determining the domain of composite functions. When you encounter a radical in any function composition, always consider what values keep the expression inside the radical non-negative.

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Most popular questions from this chapter

Will help you prepare for the material covered in the next section. $$\text { If }\left(x_{1}, y_{1}\right)=(-3,1) \text { and }\left(x_{2}, y_{2}\right)=(-2,4), \text { find } \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$$

Use a graphing utility to graph each circle whose equation is given. Use a square setting for the viewing window. $$x^{2}+y^{2}=25$$

114\. If \(f(x)=x^{2}-4\) and \(g(x)=\sqrt{x^{2}-4},\) then \((f \circ g)(x)=-x^{2}\) and \(\left(f^{\circ} g\right)(5)=-25\) 115\. There can never be two functions \(f\) and \(g\), where \(f \neq g\), for which \((f \circ g)(x)=(g \circ f)(x)\) 116\. If \(f(7)=5\) and \(g(4)=7,\) then \((f \circ g)(4)=35\) 117\. If \(f(x)=\sqrt{x}\) and \(g(x)=2 x-1,\) then \((f \circ g)(5)=g(2)\) 118\. Prove that if \(f\) and \(g\) are even functions, then \(f g\) is also an even function. 119\. Define two functions \(f\) and \(g\) so that \(f^{\circ} g=g \circ f\)

Solve for \(y: \quad x=\frac{5}{y}+4\)

Determine whether the graph of \(x^{2}-y^{3}=2\) is symmetric with respect to the \(y\) -axis, the \(x\) -axis, the origin, more than one of these, or none of these. (Section \(1.3,\) Examples 2 and 3)
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