Problem 43
Question
a. Find an equation for \(f^{-1}(x)\) b. Graph \(f\) and \(f^{-1}\) in the same rectangular coordinate system. c. Use interval notation to give the domain and the range of \(f\) and \(f^{-1}\) $$f(x)=(x-1)^{2}, x \leq 1$$
Step-by-Step Solution
Verified Answer
The inverse function of \(f(x)\) is \(f^{-1}(x)= 1-\sqrt{x}\, where \, x\geq 0.\). The graphs of both functions are parabola and decreasing square root function respectively. The domain and range of \(f(x)\) are \([- \infty ,1]\) and \([0,+\infty)\) respectively, whereas the domain and range of \(f^{-1}(x)\) are \([0,+\infty)\) and \((-\infty,1]\) respectively.
1Step 1: Find the inverse function
Let \(y = f(x)\), then \(y = (x-1)^{2}\). The inverse of \(f(x)\) can be found by interchanging \(x\) and \(y\), then solving for \(y\). Thus we have: \(x = (y-1)^2\). Taking the square root of both sides, we get \(y=1-\sqrt{x}\). Thus, the inverse function \(f^{-1}(x)= 1-\sqrt{x}\, where \, x\geq 0.\)
2Step 2: Graph the functions
The function \(f(x)=(x-1)^{2}\) is a parabola shifted one unit to the right. Because \(x \leq 1\), the graph would be the left half of the parabola. Therefore, the vertex is at (1,0), and it opens up. The inverse function \(f^{-1}(x)= 1-\sqrt{x}\) is a decreasing square root function shifted up by one unit. Since \(x \geq 0\), the graph includes all values of \(x\) from 0 to infinity.
3Step 3: Determine the domains and ranges
The domain is all allowable \(x\)-values while the range is all possible \(y\)-values. For \(f(x)=(x-1)^2\), the domain is \(x \leq 1\), expressed in interval notation as \([- \infty ,1]\) and the range is \([0,+\infty)\) since a squared real number is always nonnegative. For \(f^{-1}(x)= 1-\sqrt{x}\), the domain is \(x \geq 0\), or \([0,+\infty)\), while the range is all \(y\) less than or equal to 1, or \((-\infty,1]\).
Key Concepts
Finding Inverse FunctionsGraphing Functions and InversesDomain and Range in Interval Notation
Finding Inverse Functions
Understanding how to find an inverse function is crucial in precalculus as it lays the groundwork for solving a variety of real-world problems. An inverse function, denoted as \(f^{-1}(x)\), effectively reverses the action of the original function \(f(x)\).
The process begins by replacing \(f(x)\) with \(y\), which allows us to work with the equation in a more familiar way. Then, we swap the positions of \(x\) and \(y\) in the equation and solve for \(y\) again to determine the inverse function.
This process often involves steps such as factoring, applying the quadratic formula, or taking roots. In our example, we found the inverse of \(f(x)=(x-1)^2\) by first interchanging \(x\) and \(y\) resulting in \(x = (y-1)^2\) and then isolating \(y\), which gives us \(f^{-1}(x) = 1 - \sqrt{x}\), defined only for \(x \geq 0\). The restriction \(x \geq 0\) is due to the fact that we cannot take the square root of a negative number in the set of real numbers.
When finding inverses, it's essential to remember that not all functions have inverses that are also functions, as they must pass the 'horizontal line test'—meaning that each \(x\) value should map to only one \(y\) value in the inverse to maintain function status.
The process begins by replacing \(f(x)\) with \(y\), which allows us to work with the equation in a more familiar way. Then, we swap the positions of \(x\) and \(y\) in the equation and solve for \(y\) again to determine the inverse function.
This process often involves steps such as factoring, applying the quadratic formula, or taking roots. In our example, we found the inverse of \(f(x)=(x-1)^2\) by first interchanging \(x\) and \(y\) resulting in \(x = (y-1)^2\) and then isolating \(y\), which gives us \(f^{-1}(x) = 1 - \sqrt{x}\), defined only for \(x \geq 0\). The restriction \(x \geq 0\) is due to the fact that we cannot take the square root of a negative number in the set of real numbers.
When finding inverses, it's essential to remember that not all functions have inverses that are also functions, as they must pass the 'horizontal line test'—meaning that each \(x\) value should map to only one \(y\) value in the inverse to maintain function status.
Graphing Functions and Inverses
Visualizing functions and their inverses on a graph can provide significant insights into their behavior and relationship to one another. To effectively graph a function and its inverse, one should understand the symmetrical nature between them—they are reflections over the line \(y=x\).
Graphing starts by identifying key characteristics of the function such as intercepts, vertex, and asymptotes if applicable. In the given problem, \(f(x)=(x-1)^2\) is graphed as the left half of the parabola since \(x \leq 1\), showing where the function is defined. The graph of the inverse function, \(f^{-1}(x) = 1 - \sqrt{x}\), will be a reflection of the graph of \(f(x)\) over the line \(y=x\). Since \(f^{-1}(x)\) is defined for \(x \geq 0\), its graph extends from the point (0,1) downward to the right, representing a decreasing function. To ensure clarity in your graphs, always plot both functions on the same axes and use dashed lines to help identify the line of reflection \(y=x\).
Graphing starts by identifying key characteristics of the function such as intercepts, vertex, and asymptotes if applicable. In the given problem, \(f(x)=(x-1)^2\) is graphed as the left half of the parabola since \(x \leq 1\), showing where the function is defined. The graph of the inverse function, \(f^{-1}(x) = 1 - \sqrt{x}\), will be a reflection of the graph of \(f(x)\) over the line \(y=x\). Since \(f^{-1}(x)\) is defined for \(x \geq 0\), its graph extends from the point (0,1) downward to the right, representing a decreasing function. To ensure clarity in your graphs, always plot both functions on the same axes and use dashed lines to help identify the line of reflection \(y=x\).
Domain and Range in Interval Notation
The terms 'domain' and 'range' are foundational concepts in any function analysis. The domain refers to the set of all possible input values \(x\), and the range is the set of all possible output values \(y\). Expressing these sets in interval notation offers a concise and accurate way of describing these sets.
Interval notation incorporates brackets and parentheses to denote closed or open intervals. A closed interval, using square brackets like \([a, b]\), includes the endpoints \(a\) and \(b\), while an open interval, using parentheses like \((a, b)\), does not. For example, the domain of the function \(f(x)=(x-1)^2\) when \(x \leq 1\) is represented as \([-fty, 1]\), stating that the function includes all numbers up to and including 1. Conversely, the range of \(f(x)\) is \([0, +\infty)\), indicating that the function's output starts at 0 and includes all positive numbers.
Clear understanding and accurate use of interval notation are vital for correctly communicating domains and ranges, thus avoiding potential confusion or errors in analysis and problem-solving.
Interval notation incorporates brackets and parentheses to denote closed or open intervals. A closed interval, using square brackets like \([a, b]\), includes the endpoints \(a\) and \(b\), while an open interval, using parentheses like \((a, b)\), does not. For example, the domain of the function \(f(x)=(x-1)^2\) when \(x \leq 1\) is represented as \([-fty, 1]\), stating that the function includes all numbers up to and including 1. Conversely, the range of \(f(x)\) is \([0, +\infty)\), indicating that the function's output starts at 0 and includes all positive numbers.
Clear understanding and accurate use of interval notation are vital for correctly communicating domains and ranges, thus avoiding potential confusion or errors in analysis and problem-solving.
Other exercises in this chapter
Problem 42
Graph the given functions, \(f\) and \(g,\) in the same rectangular coordinate system. Select integers for \(x,\) starting with -2 and ending with \(2 .\) Once
View solution Problem 42
Give the slope and \(y\)-intercept of each line whose equation is given. Then graph the linear function. $$f(x)=-3 x+2$$
View solution Problem 43
Find \(f+g, f-g,\) fg, and \(\frac{f}{x}\). Determine the domain for each function. $$f(x)=\frac{5 x+1}{x^{2}-9}, g(x)=\frac{4 x-2}{x^{2}-9}$$
View solution Problem 43
Determine whether each function is even, odd, or neither. Then determine whether the function's graph is symmetric with respect to the \(y\) -axis, the origin,
View solution