Problem 6
Question
Let \(f: R \rightarrow R\) be a function defined by, \(f(x)=\frac{x^{2}-8}{x^{2}+2}\), then \(f\) is (A) one-one but not onto (B) one-one and onto (C) onto but not one-one (D) neither one-one nor onto
Step-by-Step Solution
Verified Answer
(D) neither one-one nor onto
1Step 1: Determine the nature of the function
First, let's understand what makes a function one-one and onto. A function is one-one (injective) if every element of the range is mapped by at most one element of the domain. It is onto (surjective) if every element in the codomain is an image of at least one element from the domain.For the given function \(f(x)=\frac{x^{2}-8}{x^{2}+2}\), both the numerator and denominator involve \(x^2\), which shows the function is symmetric with respect to \(y\)-axis. Thus, it cannot be injective (one-one), as multiple \(x\) values will produce the same \(f(x)\); for example, \(f(a) = f(-a)\) for any real number \(a\).
2Step 2: Check onto property
To determine if the function is onto, consider the range it covers. The expression \(f(x)\) appears as a rational function crossing horizontally all values between the limits reached as \(x\) tends to infinity or zero. Thus, we need to analyze the behavior for extreme \(x\) values:- As \(x \rightarrow \infty\), \(f(x) \rightarrow 1\).- As \(x \rightarrow -\infty\), \(f(x) \rightarrow 1\).- Evaluate \(f(0) = \frac{-8}{2} = -4\).By examining the horizontal asymptote and additional values, we find it not entirely mapped to all real numbers because the function will only approach 1, never exceeding or reaching all negative values, suggesting it is **not onto** as it misses some positive reals up to 1.
3Step 3: Conclude with function properties
Based on our analysis, the function \(f(x)=\frac{x^{2}-8}{x^{2}+2}\) is **neither one-one nor onto** as it is symmetric with respect to different inputs providing the same output, not fitting the one-one criterion, and due to limitations on the range as it never reaches some real numbers exceeding 1.
Key Concepts
One-One Function (Injective)Onto Function (Surjective)Injective and Surjective
One-One Function (Injective)
A one-one function, also known as an injective function, is a function where every element in the domain is mapped to a unique element in the range. This means no two different elements in the domain are mapped to the same element in the range. To determine if a function is one-one, you have to check if different inputs always result in different outputs.
In our original exercise, the function given is symmetric, meaning it produces the same output for corresponding positive and negative inputs. For instance, if you input a number and its negative counterpart, you will get the same result. This is because the numerator and denominator in the function are even powers of x, specifically squares. Therefore, the function is not one-one because it fails the injective test, meaning it cannot create only unique outputs for each input.
Here is a simple rule to remember:
In our original exercise, the function given is symmetric, meaning it produces the same output for corresponding positive and negative inputs. For instance, if you input a number and its negative counterpart, you will get the same result. This is because the numerator and denominator in the function are even powers of x, specifically squares. Therefore, the function is not one-one because it fails the injective test, meaning it cannot create only unique outputs for each input.
Here is a simple rule to remember:
- If a horizontal line intersects the graph of the function at more than one point, the function is not one-one.
Onto Function (Surjective)
An onto function, or surjective function, is one where every possible element of the range corresponds to an element of the domain. This means every value in the target set or codomain must have at least one pre-image in the domain. To determine if a function is onto, assess whether all potential outputs (range) are achievable from the inputs (domain).
For the function in the exercise, consider the expression: \[ f(x) = \frac{x^{2} - 8}{x^{2} + 2} \]To verify if it is onto, analyze the range of values that f(x) can take. As observed, the function approaches 1 as x tends to ±∞, but it starts from -4 when x is 0, never reaching or exceeding some positive values up to 1. This indicates that f(x) does not cover all real numbers because the approached value may not be realized. Thus, not all real numbers are output possibilities of the function.
For the function in the exercise, consider the expression: \[ f(x) = \frac{x^{2} - 8}{x^{2} + 2} \]To verify if it is onto, analyze the range of values that f(x) can take. As observed, the function approaches 1 as x tends to ±∞, but it starts from -4 when x is 0, never reaching or exceeding some positive values up to 1. This indicates that f(x) does not cover all real numbers because the approached value may not be realized. Thus, not all real numbers are output possibilities of the function.
- If a horizontal line cannot be drawn through every possible value on the y-axis to intersect the graph of the function, then it is not onto.
Injective and Surjective
Functions can be both injective and surjective, known as bijective functions. If a function is bijective, it means it is a perfect one-to-one correspondence between all elements in the domain and the range. Each element of the domain maps to a distinct and unique element in the range, and vice versa.
However, as you can see, not all functions are bijective. For example, in our exercise, the function is neither injective because of the symmetric inputs yielding equal outputs, nor surjective due to the limitations in covering the range of real numbers. Bijectivity requires that both the injective and surjective properties hold simultaneously.
However, as you can see, not all functions are bijective. For example, in our exercise, the function is neither injective because of the symmetric inputs yielding equal outputs, nor surjective due to the limitations in covering the range of real numbers. Bijectivity requires that both the injective and surjective properties hold simultaneously.
- A bijective function always has an inverse function, meaning you can "reverse" the mapping.
- Checking for both injective (one-one) and surjective (onto) properties is crucial to conclude that a function is bijective.
Other exercises in this chapter
Problem 4
Which of the following functions is are injective \(\mathrm{map}(\mathrm{s}) ?\) (A) \(f(x)=x^{2}+2, x \in(-\infty, \infty)\) (B) \(f(x)=|x+2|, x \in[-2, \infty
View solution Problem 5
The graph of the function \(\cos x \cos (x+2)-\cos ^{2}(x+1)\) is (A) a straight line passing through \(\left(0,-\sin ^{2} 1\right)\) with slope 2 (B) a straigh
View solution Problem 7
If \(f(x)=64 x^{3}+\frac{1}{x^{3}}\) and \(a, b\) are the roots of \(4 x+\frac{1}{x}=3\), then (A) \(f(a)=12\) (B) \(f(b)=11\) (C) \(f(a)=f(b)\) (D) None of the
View solution Problem 8
If the functions \(f, g, h\) are defined from the set of real numbers \(R\) to \(R\) such that \(f(x)=x^{2}-1, g(x)=\sqrt{x^{2}+1}, h(x)=\left\\{\begin{array}{l
View solution