Problem 149
Question
Assertion: Suppose, \(f(x)=(x+1)^{2}\) for \(x \geq-1\). If \(g(x)\) is the function whose graph is the reflection of the graph of \(f(x)\) with respect to the line \(y=x\), then \(g(x)=\) \(\sqrt{x}-1, x \geq 0 .\) Reason: \(g(x)\) is the inverse of \(f(x)\)
Step-by-Step Solution
Verified Answer
Yes, the assertion is true. \(g(x) = \sqrt{x} - 1\) for \(x \geq 0\) is the inverse of \(f(x) = (x+1)^2\).
1Step 1: Understand the Problem
We are given a function \(f(x) = (x+1)^2\) for \(x \geq -1\), and we need to find the function \(g(x)\) such that it is a reflection of \(f(x)\) over the line \(y = x\). The reflected graph represents the inverse of \(f(x)\).
2Step 2: Determine the Condition for Inverse
To find \(g(x)\) as the inverse of \(f(x)\), \(f(x)\) should be one-to-one on the domain \(x \geq -1\). Since the function \(f(x) = (x+1)^2\) is monotonic and increasing on this interval, it is invertible.
3Step 3: Find the Inverse Function
To find the inverse function \(g(x)\), assume \(y = f(x)\), thus \(y = (x+1)^2\). Solve for \(x\):\[ \sqrt{y} = x + 1 \]Thus,\[ x = \sqrt{y} - 1 \]Replace \(y\) with \(x\) to get the inverse function. Therefore, the inverse is \(g(x) = \sqrt{x} - 1\) for \(x \geq 0\).
4Step 4: Verify the Domain
The domain of the inverse function \(g(x) = \sqrt{x} - 1\) is determined by the range of \(f(x)\). The range of \(f(x) = (x+1)^2\) for \(x \geq -1\) is \([0, \infty)\), which confirms the domain of \(g(x)\) as \(x \geq 0\).
Key Concepts
Reflection Over Line y=xOne-to-One FunctionDomain and Range
Reflection Over Line y=x
When we talk about reflection over the line \( y = x \), we are essentially discussing how to create the mirror image of a function's graph across this diagonal line. Imagine holding a mirror along the line \( y = x \). Each point \((a, b)\) on the original function \( f(x) \) gets transformed to \((b, a)\) on the reflected graph. Therefore, reflecting a graph over the line \( y = x \) gives us the graph of the inverse function.
For example, if you're given a function like \( f(x) = (x+1)^2 \) for \( x \ge -1 \), the reflection will swap the \( x \) and \( y \) values, resulting in \( g(x) = \sqrt{x} - 1 \) for \( x \ge 0 \).
This method visually and algebraically gives us the inverse function, confirming the special relationship between reflection and inversion.
For example, if you're given a function like \( f(x) = (x+1)^2 \) for \( x \ge -1 \), the reflection will swap the \( x \) and \( y \) values, resulting in \( g(x) = \sqrt{x} - 1 \) for \( x \ge 0 \).
This method visually and algebraically gives us the inverse function, confirming the special relationship between reflection and inversion.
One-to-One Function
A one-to-one function ensures that each input has a unique output, and no output is repeated. This uniqueness is crucial when determining if a function has an inverse. Without one-to-one correspondence, a function's inverse wouldn't be valid because the inverse wouldn't pass the vertical line test. It must also pass the horizontal line test. This means that no horizontal line should intersect the graph of a function more than once.
Consider the function \( f(x) = (x+1)^2 \) where \( x \ge -1 \). This function is monotonic and increasing, which implies it is one-to-one within the specified domain.
This property also means the inverse \( g(x) = \sqrt{x} - 1 \), for \( x \ge 0 \), is valid and defined for the domain of its reverse function because the horizontal line test is satisfied. This ensures that the inverse is indeed a reflection across the line \( y = x \) without ambiguity.
Consider the function \( f(x) = (x+1)^2 \) where \( x \ge -1 \). This function is monotonic and increasing, which implies it is one-to-one within the specified domain.
This property also means the inverse \( g(x) = \sqrt{x} - 1 \), for \( x \ge 0 \), is valid and defined for the domain of its reverse function because the horizontal line test is satisfied. This ensures that the inverse is indeed a reflection across the line \( y = x \) without ambiguity.
Domain and Range
Understanding the domain and the range is crucial for inverse functions. The domain is the set of input values for which a function is defined, whereas the range is the set of possible output values. The trick here is: when you find the inverse, the domain and range swap places.
In the example of \( f(x) = (x+1)^2 \) for \( x \ge -1 \), the domain is \( x \ge -1 \). The resulting output or range is \( [0, \infty) \).
However, for the inverse function \( g(x) = \sqrt{x} - 1 \), the range of \( f \) becomes the domain of \( g \), so the domain becomes \( x \ge 0 \). Similarly, the inverse's range is symmetric, reflecting the swapped domain of \( f \) which would be \( y \ge -1 \).
This swap is essential for correctly graphing and understanding inverse functions and their behavior across their respective domains and ranges.
In the example of \( f(x) = (x+1)^2 \) for \( x \ge -1 \), the domain is \( x \ge -1 \). The resulting output or range is \( [0, \infty) \).
However, for the inverse function \( g(x) = \sqrt{x} - 1 \), the range of \( f \) becomes the domain of \( g \), so the domain becomes \( x \ge 0 \). Similarly, the inverse's range is symmetric, reflecting the swapped domain of \( f \) which would be \( y \ge -1 \).
This swap is essential for correctly graphing and understanding inverse functions and their behavior across their respective domains and ranges.
Other exercises in this chapter
Problem 143
Column-I I. Range of the function \(f(x)=\sqrt{3 x^{2}-4 x+5}\) II. Range of the function \(f(x)=\log _{\varepsilon}\left(3 x^{2}-4 x+5\right)\) III. The value
View solution Problem 146
Assertion: If \(f(x)=\frac{a^{x}}{a^{x}+\sqrt{a}}(a>0)\), then \(\sum_{r=1}^{2 n-1} 2 f\left(\frac{r}{2 n}\right)=2 n-1\) Reason: \(f(x)+f(1-x)=1 \forall x\)
View solution Problem 150
The period of \(\sin ^{2} \theta\) is : (A) \(\pi^{2}\) (B) \(\pi\) (C) \(2 \pi\) (D) \(\pi 2\)
View solution Problem 151
The domain of \(\sin ^{-1}\left[\log _{3}(x / 3)\right]\) is: (A) \([1,9]\) (B) \([-1,9]\) (C) \([-9,1]\) (D) \([-9,-1]\)
View solution