Problem 65
Question
Find the domain of each function. $$ f(x)=\frac{1}{\sqrt{x-3}} $$
Step-by-Step Solution
Verified Answer
The domain of the function \(f(x) = \frac{1}{\sqrt{x-3}}\) is \(x > 3\).
1Step 1: Identify the Original Function
Identify that the given function \(f(x)=\frac{1}{\sqrt{x-3}}\) is a combination of a reciprocal and a square root function.
2Step 2: Set the Function Inside the Radical Greater than Zero
Set the inside function (the one inside the square root) which is \(x-3\) greater than zero. The reason we do this is because we cannot take the square root of a negative number within the real number system, thus for the square root function to be defined, \(x-3\) must be more than or equal to zero. So, solve the inequality \(x-3 > 0\).
3Step 3: Solve the Inequality
Solving the inequality \(x-3 > 0\), we will get \(x > 3\). However, we have to be careful because 3 is not included in the solution because substituting 3 results in division by zero which is undefined.
4Step 4: State the Domain
The solution to the inequality give us the domain of the function. So, the domain of the function \(f(x)=\frac{1}{\sqrt{x-3}}\) is \(x > 3\). That means the function is defined for all real numbers greater than 3.
Key Concepts
Reciprocal FunctionSquare Root FunctionInequality SolvingReal Numbers
Reciprocal Function
A reciprocal function is a function that involves the division of 1 by a variable. In mathematical terms, it is often expressed as \( y = \frac{1}{x} \). This basic form tells us that as \( x \) approaches zero, the function's value grows infinitely positive or negative, leading to a graph where \( x = 0 \) is a vertical asymptote.
These characteristics make the reciprocal function interesting:
These characteristics make the reciprocal function interesting:
- The function is undefined at \( x = 0 \), resulting in a vertical asymptote.
- When positive numbers are used, the function yields positive results.
- Negative inputs result in negative outputs, reflecting symmetry with respect to the origin.
Square Root Function
The square root function is defined as \( y = \sqrt{x} \). This function maps a number to its square root, and it is only defined for non-negative inputs within the real number system.
- The reason behind this restriction is that square roots of negative numbers are not real.
- The graph of a simple square root function is a curve that starts at the origin (0,0) and rises slowly to the right.
- Any output is a non-negative real number.
Inequality Solving
In the context of finding the function's domain, solving inequalities helps identify where the function is defined.
For instance, in our exercise, we have \( x-3 > 0 \) to ensure the function is inside the square root is defined in the realm of real numbers.
For instance, in our exercise, we have \( x-3 > 0 \) to ensure the function is inside the square root is defined in the realm of real numbers.
- Simplifying the inequality \( x-3 > 0 \) involves isolating \( x \).
- This translates to \( x > 3 \), which means only numbers greater than 3 can be inputs to the function.
Real Numbers
Real numbers encompass all numbers along the continuous number line, including both rational and irrational numbers.
- Rational numbers can be expressed as fractions, like \( \frac{1}{2} \), while irrational numbers cannot, such as \( \sqrt{2} \).
- The domain of most functions is limited to real numbers, for predictability and defined results.
- When excluding certain values (like in our exercise), those values are often points where functions become undefined or imaginary.
Other exercises in this chapter
Problem 64
Determine whether each function is even, odd, or neither. $$g(x)=x^{2}-x$$
View solution Problem 65
a. Use a graphing utility to graph \(f(x)=x^{2}+1\) b. Graph \(f(x)=x^{2}+1,\) and \(g(x)=f\left(\frac{1}{2} x\right),\) and \(h(x)=f\left(\frac{1}{4} x\right)\
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Use the given conditions to write an equation for each line in point-slope form and slope-intercept form. Passing through \((-2,2)\) and parallel to the line wh
View solution Problem 65
If \(f(x)=3 x\) and \(g(x)=x+5,\) find \((f \circ g)^{-1}(x)\) and \(\left(g^{-1} \circ f^{-1}\right)(x)\)
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