Problem 21
Question
Given \(f(x)=\sqrt{2-4 x}\) and \(g(x)=-\frac{3}{x},\) fi \(\mathrm{d}\) the following: a. \((g \circ f)(x)\) b. the domain of \((g \circ f)(x)\) in interval notation
Step-by-Step Solution
Verified Answer
a. \((g \circ f)(x) = -\frac{3}{\sqrt{2-4x}}\); b. Domain: \((-\infty, \frac{1}{2})\).
1Step 1: Understanding Composition of Functions
To find \((g \circ f)(x)\), we need to substitute \(f(x)\) into \(g(x)\). This means wherever there is an \(x\) in \(g(x)\), we replace it with \(f(x)\). Since \(g(x) = -\frac{3}{x}\) and \(f(x) = \sqrt{2-4x}\), we find \((g \circ f)(x)\) as \(g(f(x)) = -\frac{3}{\sqrt{2-4x}}\).
2Step 2: Compute the Expressed Formula
Using the substitution, \(g(f(x)) = -\frac{3}{\sqrt{2-4x}}\), we have now found the explicit expression for \((g \circ f)(x)\).
3Step 3: Determine the Domain of \(f(x)\)
\(f(x) = \sqrt{2-4x}\) is defined only if the expression under the square root is non-negative. Thus, we require \(2-4x \geq 0\). Solving \(2-4x \geq 0\) gives \(-4x \geq -2\), which simplifies to \(x \leq \frac{1}{2}\). Consequently, the domain of \(f(x)\) is \((-\infty, \frac{1}{2}]\).
4Step 4: Determine the Domain of \(g(f(x))\)
Since \(g(f(x)) = -\frac{3}{\sqrt{2-4x}}\), we must ensure that the expression in the denominator, \(\sqrt{2-4x}\), is non-zero (i.e., \(\sqrt{2-4x} eq 0\)) in addition to being defined. Therefore, \(x < \frac{1}{2}\). Combining this with the domain of \(f(x)\) from Step 3, the domain of \((g \circ f)(x)\) is \((-\infty, \frac{1}{2})\).
5Step 5: Write Domain in Interval Notation
From Step 4, the domain of \((g \circ f)(x)\) in interval notation is \((-\infty, \frac{1}{2})\).
Key Concepts
Domain of a FunctionInterval NotationSquare Root Function
Domain of a Function
The domain of a function is the complete set of possible values of the independent variable, usually denoted as "x", for which the function is defined. To find the domain, you need to consider conditions such as prohibiting division by zero or ensuring that expressions under a square root symbol are non-negative.
In our exercise, we have the function \(f(x) = \sqrt{2-4x}\). The domain requires that the expression under the square root is non-negative. This leads to solving the inequality \(2-4x \geq 0\). Simplifying gives \(x \leq \frac{1}{2}\), meaning \(f(x)\) is defined for all \(x\) values up to \(\frac{1}{2}\).
When finding the domain of a composition of functions, like \((g \circ f)(x)\), consider the domain restrictions of each individual function and the composition. Since \(g(x) = -\frac{3}{x}\), the expression is only defined when the denominator, \(x\), is non-zero. For \(g(f(x)) = -\frac{3}{\sqrt{2-4x}}\), ensure \(\sqrt{2-4x} eq 0\), leading to \(x < \frac{1}{2}\). The resulting combined domain is \(( - \infty, \frac{1}{2} )\).
In our exercise, we have the function \(f(x) = \sqrt{2-4x}\). The domain requires that the expression under the square root is non-negative. This leads to solving the inequality \(2-4x \geq 0\). Simplifying gives \(x \leq \frac{1}{2}\), meaning \(f(x)\) is defined for all \(x\) values up to \(\frac{1}{2}\).
When finding the domain of a composition of functions, like \((g \circ f)(x)\), consider the domain restrictions of each individual function and the composition. Since \(g(x) = -\frac{3}{x}\), the expression is only defined when the denominator, \(x\), is non-zero. For \(g(f(x)) = -\frac{3}{\sqrt{2-4x}}\), ensure \(\sqrt{2-4x} eq 0\), leading to \(x < \frac{1}{2}\). The resulting combined domain is \(( - \infty, \frac{1}{2} )\).
Interval Notation
Interval notation is a mathematical notation used to describe the set of all numbers between a pair of endpoints. It is very useful when identifying the domain of a function.
When writing interval notation, a closed interval uses brackets \([a, b]\), indicating inclusivity of the endpoints \(a\) and \(b\). An open interval uses parentheses \((a, b)\), excluding the endpoints from the set.
For the domain of \(f(x)=\sqrt{2-4x}\), which is \(x \leq \frac{1}{2}\), the interval notation is \(( - \infty, \frac{1}{2} ]\). This includes every real number less than or equal to \( \frac{1}{2}\). However, in the composition \(g(f(x))\), the interval does not include \(x = \frac{1}{2}\), changing it to \(( - \infty, \frac{1}{2} )\). This emphasizes the importance of specifying endpoints in interval notation.
When writing interval notation, a closed interval uses brackets \([a, b]\), indicating inclusivity of the endpoints \(a\) and \(b\). An open interval uses parentheses \((a, b)\), excluding the endpoints from the set.
For the domain of \(f(x)=\sqrt{2-4x}\), which is \(x \leq \frac{1}{2}\), the interval notation is \(( - \infty, \frac{1}{2} ]\). This includes every real number less than or equal to \( \frac{1}{2}\). However, in the composition \(g(f(x))\), the interval does not include \(x = \frac{1}{2}\), changing it to \(( - \infty, \frac{1}{2} )\). This emphasizes the importance of specifying endpoints in interval notation.
Square Root Function
The square root function is commonly expressed as \(\sqrt{x}\). It is only defined for non-negative values of \(x\) due to the nature of square roots. The value of \(\sqrt{x}\) is the number that, when multiplied by itself, gives \(x\).
In the context of our exercise, \(f(x) = \sqrt{2-4x}\), the expression under the square root must be non-negative. This condition \(2-4x \geq 0\) ensures the square root function is defined, resulting in \(x \leq \frac{1}{2}\).
The characteristic shape of the square root function is a curve that starts from a point and moves upward and to the right, representing an increase as \(x\) becomes larger within the domain. This understanding helps in analyzing how modifications in the argument, like \(2-4x\), shift the domain and range of the function.
In the context of our exercise, \(f(x) = \sqrt{2-4x}\), the expression under the square root must be non-negative. This condition \(2-4x \geq 0\) ensures the square root function is defined, resulting in \(x \leq \frac{1}{2}\).
The characteristic shape of the square root function is a curve that starts from a point and moves upward and to the right, representing an increase as \(x\) becomes larger within the domain. This understanding helps in analyzing how modifications in the argument, like \(2-4x\), shift the domain and range of the function.
Other exercises in this chapter
Problem 21
Given \(f(x)=\sqrt{2-4 x}\) and \(g(x)=-\frac{3}{x},\) find the following: a. \((g \circ f)(x)\) b. the domain of \((g \circ f)(x)\) in interval notation
View solution Problem 21
Determine the interval(s) on which the function is increasing and decreasing. $$g(x)=5(x+3)^{2}-2$$
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For the following exercises, determine whether the relation represents \(y\) as a function of \(x\). $$ y=x^{3} $$
View solution Problem 21
For the following exercises, find the domain of each function using interval notation. $$ f(x)=\frac{2 x+1}{\sqrt{5-x}} $$
View solution