/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 You are given a pair of function... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given a pair of functions, \(f\) and \(g .\) In each case, use your grapher to estimate the domain of \((g \circ f)(x)\). Confirm analytically. $$ f(x)=\sqrt{x-5}, g(x)=x^{2} $$

Short Answer

Expert verified
The domain of \((g \circ f)(x)\) is \([5, \infty)\)."}

Step by step solution

01

Determine the Domain of f

The function \( f(x) = \sqrt{x-5} \) requires that the expression inside the square root be non-negative to be defined. Therefore, the domain of \( f \) is determined by \( x-5 \geq 0 \Rightarrow x \geq 5 \). Thus, the domain of \( f(x) \) is \([5, \infty)\).
02

Determine the Domain of g

The function \( g(x) = x^2 \) is a polynomial, which is defined for all real numbers. Therefore, the domain of \( g(x) \) is \((\infty, \infty)\).
03

Determine the Domain of g(f(x))

The composite function \( (g \circ f)(x) = g(f(x)) = (\sqrt{x-5})^2 \) involves squaring the output of \( f(x) \). Since the output of \( f(x) \) is defined from Step 1, the domain of \( g(f(x)) \) is the same as the domain of \( f(x) \), which is \([5, \infty)\).
04

Verify with a Graph

Use a graphing calculator or software to graph \( (g \circ f)(x) = (\sqrt{x-5})^2 \) and observe the graph. The graph will show that \( (g \circ f)(x) \) starts at \( x = 5 \) and extends to the right, confirming that the domain is \([5, \infty)\).

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Key Concepts

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

Composite Functions
A composite function is formed when one function is applied, and then another function is applied to the result of the first one. For example, if we have functions \( f \) and \( g \), the composite function \( (g \circ f)(x) \) represents applying function \( f \) to \( x \) first, and then applying function \( g \) to the result of \( f(x) \).
This can be graphically represented, where the output of \( f(x) \) becomes the input for \( g(x) \). For our exercise, where \( f(x) = \sqrt{x-5} \) and \( g(x) = x^2 \), forming \( (g \circ f)(x) \) involves substituting the output of \( f \) into \( g \).
  • First, calculate \( f(x) \) which gives \( \sqrt{x-5} \).
  • Then, substitute that into \( g \), giving \( (g \circ f)(x) = (\sqrt{x-5})^2 \).
It is crucial to understand how the domain of the composite function is influenced. The composite's domain is based on where \( f(x) \) is defined, considering that the output of \( f(x) \) becomes the input for \( g(x) \).
Always ensure the initial input for \( f(x) \) matches its domain requirements, and subsequently, that \( g(x) \) can process what \( f \) produces.
Square Root Function
The square root function, like \( f(x) = \sqrt{x-5} \), requires cautious consideration of its domain. This is because the expression inside the root must be non-negative (i.e., zero or positive).
This requirement avoids producing any complex numbers, as real-valued square roots for negative numbers do not exist.
In the function \( \sqrt{x-5} \), the expression inside the square root, \( x-5 \), must satisfy \( x-5 \geq 0 \).
Solving this inequality, we find the domain as \( x \geq 5 \) or expressed in interval notation as \([5, \infty)\).
  • The square root function introduces a constraint that ensures the output remains real.
  • This domain condition also dictates subsequent operations when combined with other functions.
Understanding this is fundamental because it directly impacts the function's graph, which will only start plotting from where \( x = 5 \), progressing towards infinity on the right along the x-axis.
Polynomial Functions
Polynomial functions, such as \( g(x) = x^2 \), are quite flexible in terms of domains. This flexibility is due to polynomials being defined for all real numbers.
Such functions have no restrictions like square root or division by zero conditions, making their domain \( (-\infty, \infty) \).
This uninhibited domain means you can input any real value into \( g(x) \) and expect a real number in response.
  • In polynomial functions, degree and coefficients determine the graph's curvature and behavior.
  • Particularly, \( x^2 \) creates a parabolic shape that extends indefinitely left and right.
For composite functions, the flexibility of polynomial functions allows them to accept a variety of inputs provided they come from another function’s valid range.
In our case, \( g(x) = x^2 \) accepts outputs from \( f(x) = \sqrt{x-5} \). This feature ensures that once we have resolved any domain constraints from initial functions, the polynomial function’s scope does not further restrict the composite function's domain.

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