Problem 35
Question
Find \(h(x)=f(g(x))\) and \(j(x)=g(f(x)) .\) What are the domains of \(h\) and \(j\) ? $$ f(x)=x^{2}+9 \text { and } g(x)=\frac{1}{\sqrt{x}} $$
Step-by-Step Solution
Verified Answer
Thus, \(h(x) = \frac{1}{x} + 9\) with a domain of \((-\infty, 0) \cup (0, +\infty)\), and \(j(x) = \frac{1}{\sqrt{x^2+9}}\) with a domain of \((-\infty, +\infty)\).
1Step 1: Compute the function \(h(x) = f(g(x))\)
Substitute \(g(x)\) into the function \(f(x)\). That is \(h(x) = (g(x))^2 + 9 = (\frac{1}{\sqrt{x}})^2 + 9 = \frac{1}{x} + 9. \)
2Step 2: Compute the function \(j(x) = g(f(x))\)
Next, substitute \(f(x)\) into the function \(g(x)\). That is \(j(x) = \frac{1}{\sqrt{f(x)}} = \frac{1}{\sqrt{x^2+9}}. \)
3Step 3: Find the domain of \(h(x)\)
The domain of \(h\) is all real \(x\) for which \(h(x)\) is defined. Since \(\frac{1}{x}\) in \(h(x) = \frac{1}{x} + 9\) is not defined for \(x = 0\) , the domain of \(h(x)\) is \((-\infty, 0) \cup (0, +\infty)\).
4Step 4: Find the domain of \(j(x)\)
The domain of \(j\) is all real \(x\) for which \(j(x)\) is defined. There's no restriction on \(x\) in \(j(x) = \frac{1}{\sqrt{x^2+9}}\), hence the domain of \(j(x)\) is \((-\infty, +\infty)\).
Key Concepts
Domain of a functionFunction compositionSquare root function
Domain of a function
Understanding the domain of a function is essential because it tells us all the possible input values (x-values) that we can safely use in a function. These are the values for which the function is defined and does not result in any mathematical errors.
For example, in the function \( h(x) = \frac{1}{x} + 9 \), the term \( \frac{1}{x} \) causes a problem when \( x = 0 \), because division by zero is undefined. Thus, the domain excludes zero, yielding the domain as \( (-\infty, 0) \cup (0, +\infty) \).
In contrast, for the function \( j(x) = \frac{1}{\sqrt{x^2+9}} \), there are no such restrictions, as \( x^2 + 9 \) is always positive, and taking a square root of a positive number does not pose any problems. Hence, the domain here is all real numbers, or \( (-\infty, +\infty) \).
For example, in the function \( h(x) = \frac{1}{x} + 9 \), the term \( \frac{1}{x} \) causes a problem when \( x = 0 \), because division by zero is undefined. Thus, the domain excludes zero, yielding the domain as \( (-\infty, 0) \cup (0, +\infty) \).
In contrast, for the function \( j(x) = \frac{1}{\sqrt{x^2+9}} \), there are no such restrictions, as \( x^2 + 9 \) is always positive, and taking a square root of a positive number does not pose any problems. Hence, the domain here is all real numbers, or \( (-\infty, +\infty) \).
Function composition
Function composition involves combining two functions where the output of one function becomes the input of another. This is similar to performing an action and then applying another action to the result.
For instance, in the exercise, we combine functions \( f(x) = x^2 + 9 \) and \( g(x) = \frac{1}{\sqrt{x}} \) to form two new functions:
Similarly, for \( j(x) = g(f(x)) \), \( f(x) \) is placed into \( g(x) \), resulting in \( j(x) = \frac{1}{\sqrt{x^2+9}} \). This step-by-step substitution highlights how function composition transforms inputs through multiple stages.
For instance, in the exercise, we combine functions \( f(x) = x^2 + 9 \) and \( g(x) = \frac{1}{\sqrt{x}} \) to form two new functions:
- \( h(x) = f(g(x)) \)
- \( j(x) = g(f(x)) \)
Similarly, for \( j(x) = g(f(x)) \), \( f(x) \) is placed into \( g(x) \), resulting in \( j(x) = \frac{1}{\sqrt{x^2+9}} \). This step-by-step substitution highlights how function composition transforms inputs through multiple stages.
Square root function
The square root function, denoted as \( \sqrt{x} \), is essential to understand because it has unique properties that affect its domain and composition with other functions. It only accepts non-negative numbers (\( x \geq 0 \)), since the square root of a negative number is not a real number.
However, when the square root function is combined with other functions, like \( g(x) = \frac{1}{\sqrt{x}} \), it imposes additional conditions on the domain. The expression \( \sqrt{x} \) in the denominator implies \( x \gt 0 \) to avoid division by zero or negative numbers.
The example in the exercise showcases such combinations, where function \( g(x) = \frac{1}{\sqrt{x}} \) interacts with \( f(x) = x^2 + 9 \) when forming \( j(x) \). The expression ensures the domain conditions remain satisfied, resulting in a smooth functioning of the composite function without errors.
However, when the square root function is combined with other functions, like \( g(x) = \frac{1}{\sqrt{x}} \), it imposes additional conditions on the domain. The expression \( \sqrt{x} \) in the denominator implies \( x \gt 0 \) to avoid division by zero or negative numbers.
The example in the exercise showcases such combinations, where function \( g(x) = \frac{1}{\sqrt{x}} \) interacts with \( f(x) = x^2 + 9 \) when forming \( j(x) \). The expression ensures the domain conditions remain satisfied, resulting in a smooth functioning of the composite function without errors.
Other exercises in this chapter
Problem 33
Let \(f(x)=|x|, g(x)=\sqrt{x}\), and \(h(x)=x-2\). Find the domain for each of the following. (a) \(p(x)=h(g(h(x)))\) (b) \(q(x)=f(h(g(x)))\)
View solution Problem 34
Let \(f(x)=x^{2}+9, g(x)=\sqrt{x}\), and \(h(x)=g(f(x))\). Find the average rate of change of \(h\) over the following intervals. (a) \([-4,4]\) (b) \([0,4]\) (
View solution Problem 36
Find \(h(x)=f(g(x))\) and \(j(x)=g(f(x)) .\) What are the domains of \(h\) and \(j\) ? $$ f(x)=\frac{2}{x+2} \text { and } g(x)=x-2 $$
View solution Problem 37
Find \(h(x)=f(g(x))\) and \(j(x)=g(f(x)) .\) What are the domains of \(h\) and \(j\) ? $$ f(x)=x^{2} \text { and } g(x)=-2 x+3 $$
View solution