Problem 39
Question
Find the inverse function of \(f,(\mathbf{b})\) graph \(f\) and \(f^{-1}\) on the same set of coordinate axes, ( \(\mathbf{c} )\) describe the relationship between the graphs, and ( \(\mathbf{d} )\) state the domain and range of \(f\) and \(f^{-1} .\) \(f(x)=\sqrt{x}\)
Step-by-Step Solution
Verified Answer
The inverse function of \(f(x) = \sqrt{x}\) is \(f^{-1}(x) = x^2\). The graphs of the function and its inverse are mirror images about the line \(y = x\). The domain and range of \(f(x) = \sqrt{x}\) are \(x \geq 0\) and \(y \geq 0\) respectively. The domain and range of its inverse \(f^{-1}(x) = x^2\) are all real numbers and \(y \geq 0\) respectively.
1Step 1: Finding the Inverse Function
To find the inverse of a function we typically swap \(x\)'s with \(y\)'s and solve the equation for \(y\). The function \(f\) is given as \(f(x) = \sqrt{x}\). Rewriting it as \(y = \sqrt{x}\), swapping \(x\)'s with \(y\)'s gives us \(x = \sqrt{y}\), square each side, we get \(x^2 = y\). Therefore, the inverse function, \(f^{-1}(x) = x^2 \).
2Step 2: Graphing the Function and its Inverse
Both functions can be graphed on the coordinate axes. The function \(f(x) = \sqrt{x}\) is the upper half of the parabola and its inverse function \(f^{-1}(x) = x^2\) is a parabola.
3Step 3: Describing the Relationship Between the graphs
The graph of \(f(x) = \sqrt{x}\) and its inverse \(f^{-1}(x) = x^2 \) are mirror images of one another with respect to the line \(y = x\). This is a property of inverse functions: they are reflected in the line \(y = x\).
4Step 4: Stating the Domains and Ranges
The domain of a function is the set of all possible input values (x-values) while the range is the set of all possible output values (y-values). For the function \(f(x) = \sqrt{x}\), the domain is \(x \geq 0\), and the range is \(y \geq 0\). For \(f^{-1}(x) = x^2\), the domain is all real numbers and the range is \(y \geq 0\).
Key Concepts
Function GraphingDomain and RangeMirror Image PropertySquare Root Function
Function Graphing
Graphing functions is an essential way to visualize mathematical relationships. When we graph a function like
By seeing them together, you gain insight into how these functions behave. The function \( f(x) = \sqrt{x} \) maps inputs to outputs in a curved line that starts at the origin [ (0,0) ] and rises to the right. On the other hand, its inverse, \( f^{-1} (x) = x^2 \), sweeps symmetrically around the y-axis, from (0,0) outward
in both directions. These visuals help us understand the nature and relationship of functions and their inverses.
- Function: \( f(x) = \sqrt{x} \) is part of a parabola's upper half.
- Inverse Function: \( f^{-1}(x) = x^2 \) is the whole U-shaped parabola.
By seeing them together, you gain insight into how these functions behave. The function \( f(x) = \sqrt{x} \) maps inputs to outputs in a curved line that starts at the origin [ (0,0) ] and rises to the right. On the other hand, its inverse, \( f^{-1} (x) = x^2 \), sweeps symmetrically around the y-axis, from (0,0) outward
in both directions. These visuals help us understand the nature and relationship of functions and their inverses.
Domain and Range
Every function has a domain and a range, essential in understanding its boundaries. For our square root function,
- Domain of \( f(x) = \sqrt{x} \): This is the set of all acceptable inputs. For our function, this is \( x \geq 0 \).
- Range of \( f(x) = \sqrt{x} \): This is the set of resultant output values. So, when \( x \geq 0 \), \( y \geq 0 \) too.
- Domain of the inverse, \( f^{-1}(x) = x^2 \): This is the whole spectrum of real numbers since any real number can be squared.
- Range of \( f^{-1}(x) = x^2 \): This reflects only positive outcomes because squaring produces non-negative results.
Mirror Image Property
When we explore inverse functions, an exciting property emerges: reflection symmetry. In our case, the functions
This property is a hallmark of function and inverse pairs. If one graph can be reflected over the line \( y = x \) to coincide with the other, youcan be confident they are inverses.
This principle not only confirms that we have found the correct inverse but illustrates the nature of these interconnected graphs.
- \( f(x) = \sqrt{x} \) and its inverse, \( f^{-1}(x) = x^2 \),
This property is a hallmark of function and inverse pairs. If one graph can be reflected over the line \( y = x \) to coincide with the other, youcan be confident they are inverses.
This principle not only confirms that we have found the correct inverse but illustrates the nature of these interconnected graphs.
Square Root Function
The square root function, \( f(x) = \sqrt{x} \), holds special characteristics:
Recognizing the \( \sqrt{x} \) graph's shape helps identify its direction and boundary. And knowing it's inverse, how this standard offset of the function is hyper-important in core math concepts.
- It originates from the non-negative half of real numbers. You cannot extract a square root of a negative number in real numbers.
- Its graph begins at (0,0), curving upward steadily.
- This function only produces non-negative outputs. Thus, every input from zero onward gives a positive or zero result.
Recognizing the \( \sqrt{x} \) graph's shape helps identify its direction and boundary. And knowing it's inverse, how this standard offset of the function is hyper-important in core math concepts.
Other exercises in this chapter
Problem 39
In Exercises 37–40, find the limit. $$ \lim _{x \rightarrow 2^{-}} \ln \left[x^{2}(3-x)\right] $$
View solution Problem 39
Finding an Indefinite Integral of a Trigonometric Function In Exercises \(31-40\) , find the indefinite integral. $$ \int \frac{\sec x \tan x}{\sec x-1} d x $$
View solution Problem 40
Find the derivative of the function. \(f(t)=\arcsin t^{2}\)
View solution Problem 40
In Exercises 37–40, find any relative extrema of the function. Use a graphing utility to confirm your result. $$ h(x)=2 \tanh x-x $$
View solution