Problem 20
Question
In \(17-20 :\) a. Find the inverse of each given function. b. Describe the domain and range of each given function and its inverse in terms of the largest possible subset of the real numbers. $$ \mathrm{f}(x)=\sqrt{x} $$
Step-by-Step Solution
Verified Answer
Inverse: \( f^{-1}(x) = x^2 \); Domain: \( x \geq 0 \); Range: \( y \geq 0 \).
1Step 1: Understanding the Original Function
The given function is \( f(x) = \sqrt{x} \). This is the square root function, which originates from the non-negative real numbers. The domain of this function is \( x \geq 0 \). Since the square root yields non-negative outputs, the range of \( f(x) \) is \( y \geq 0 \).
2Step 2: Find the Inverse Function
To find the inverse function \( f^{-1}(x) \), we start by replacing \( f(x) \) with \( y \), so \( y = \sqrt{x} \). Next, we solve for \( x \) in terms of \( y \). Square both sides to get \( y^2 = x \). Thus, the inverse function is \( f^{-1}(x) = x^2 \).
3Step 3: Domain and Range of the Inverse Function
For the inverse function \( f^{-1}(x) = x^2 \), the domain is determined by the range of the original function. Since \( f(x) \) has a range of \( y \geq 0 \), the domain of \( f^{-1}(x) \) is \( x \geq 0 \). The range of \( x^2 \) is all non-negative real numbers, which matches the domain of the function \( f(x) = \sqrt{x} \).
4Step 4: Conclusion
The inverse of the function \( f(x) = \sqrt{x} \) is \( f^{-1}(x) = x^2 \). The domain and range are restricted to non-negative real numbers for both the original function and its inverse, reflecting the property that both \( \sqrt{x} \) and \( x^2 \) produce non-negative outputs when taking non-negative inputs.
Key Concepts
Domain and RangeSquare Root FunctionReal Numbers
Domain and Range
When dealing with functions, two critical concepts are domain and range. The **domain** is the set of all possible input values a function can accept. In the world of mathematics, it’s like the menu of options you can choose from. In contrast, the **range** is all possible outputs or results that come out of the function. Imagine it as the list of all dishes you might receive after making a choice from the menu.
For the function \( f(x) = \sqrt{x} \), finding the domain and range is straightforward because it is a **square root function**. The square root is defined only for non-negative real numbers. Thus, its domain is \( x \geq 0 \). This means you can only input zero and positive numbers. On the output side, because it only yields non-negative results, its range is \( y \geq 0 \).
When considering its inverse, \( f^{-1}(x) = x^2 \), these sets flip in roles. The range of \( f(x) \), which is \( y \geq 0 \), becomes the domain of \( f^{-1}(x) \). And because squaring a non-negative real number still results in a non-negative number, the range of \( f^{-1}(x) \) is also \( y \geq 0 \).
Understanding domain and range is crucial not just for grasping individual functions but for comprehending how outputs become inputs in their inverse counterparts.
For the function \( f(x) = \sqrt{x} \), finding the domain and range is straightforward because it is a **square root function**. The square root is defined only for non-negative real numbers. Thus, its domain is \( x \geq 0 \). This means you can only input zero and positive numbers. On the output side, because it only yields non-negative results, its range is \( y \geq 0 \).
When considering its inverse, \( f^{-1}(x) = x^2 \), these sets flip in roles. The range of \( f(x) \), which is \( y \geq 0 \), becomes the domain of \( f^{-1}(x) \). And because squaring a non-negative real number still results in a non-negative number, the range of \( f^{-1}(x) \) is also \( y \geq 0 \).
Understanding domain and range is crucial not just for grasping individual functions but for comprehending how outputs become inputs in their inverse counterparts.
Square Root Function
The square root function, \( f(x) = \sqrt{x} \), is a common and significant mathematical function we encounter. It fundamentally "undoes" a square. When you square a number, you multiply it by itself; taking the square root is like asking, "What number, when squared, gives me this result?"
Because squaring any real number always results in a non-negative value, the square root function is naturally defined only for non-negative numbers. This makes the square root a **one-sided function** on real numbers.
Because squaring any real number always results in a non-negative value, the square root function is naturally defined only for non-negative numbers. This makes the square root a **one-sided function** on real numbers.
- Visualization: If you imagine the graph of \( f(x) = \sqrt{x} \), it starts at the origin (0,0) and curves gently upwards to the right.
- Behavior: It grows more slowly than linear functions, meaning for each increase in \( x \), the jump in \( y \) gets smaller and smaller.
Real Numbers
The term **real numbers** refers to all the numbers on the number line, including all the rational and irrational numbers. They are the bread and butter of most mathematical problems we solve. These numbers encompass:
- Natural Numbers: Counting numbers like 1, 2, 3, etc.
- Whole Numbers: Natural numbers plus zero (0).
- Integers: Whole numbers plus their negatives (..., -2, -1, 0, 1, 2, ...).
- Rational Numbers: Numbers that can be expressed as a fraction of two integers, such as \( \frac{1}{2} \) or \(-\frac{3}{4} \).
- Irrational Numbers: Numbers that cannot be written as a simple fraction, like \( \pi \) or \( \sqrt{2} \).
Other exercises in this chapter
Problem 19
In \(13-20\) : a. Graph each function. b. Is the function a direct variation? \(c\) . Is the function one-to-one? \(\frac{y}{x}=2\)
View solution Problem 19
In \(12-23,\) each set is a function from set \(A\) to set \(B .\) a. What is the largest subset of the real numbers that can be set \(A\) , the domain of the g
View solution Problem 20
In \(20-27\) : a. Write each equation in center-radius form. b. Find the coordinates of the center. . Find the radius of the circle. $$ x^{2}+y^{2}-25=0 $$
View solution Problem 20
A polynomial function of degree three, \(\mathrm{p}(x),\) intersects the \(x\) -axis at \((-4,0),(-2,0),\) and \((3,0)\) and intersects the \(y\) -axis at \((0,
View solution