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If f is a function defined from R to R, is given by f(x) = x2 then f -1(x) is given by?
- a)1/(3x-5)
- b)(x+5)/3
- c)does not exist since it is not a bijection
- d)none of the mentioned
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If f is a function defined from R to R, is given by f(x) = x2then f-1(...
Explanation:
In order for a function to have an inverse, it must be a bijection, which means it must be both injective (one-to-one) and surjective (onto). Let's analyze the given function to determine if it is a bijection.
Injectivity:
To check if a function is injective, we need to ensure that different inputs map to different outputs. In this case, if we have two different inputs x1 and x2 such that f(x1) = f(x2), then x1 = x2 must hold for the function to be injective.
Given f(x) = x^2, let's assume f(x1) = f(x2):
x1^2 = x2^2
Taking the square root of both sides, we get:
|x1| = |x2|
Now, we can see that for any positive value of x, there are two different inputs (x and -x) that give the same output. Therefore, the function is not injective.
Surjectivity:
To check if a function is surjective, we need to ensure that every element in the codomain has a pre-image in the domain. In this case, the function f(x) = x^2 maps every real number x to a non-negative real number.
However, there are non-negative real numbers that do not have a real square root. For example, there is no real number x such that x^2 = -1. Therefore, the function is not surjective.
Since the function is neither injective nor surjective, it does not have an inverse. Hence, the correct answer is option C: does not exist since it is not a bijection.
In order for a function to have an inverse, it must be a bijection, which means it must be both injective (one-to-one) and surjective (onto). Let's analyze the given function to determine if it is a bijection.
Injectivity:
To check if a function is injective, we need to ensure that different inputs map to different outputs. In this case, if we have two different inputs x1 and x2 such that f(x1) = f(x2), then x1 = x2 must hold for the function to be injective.
Given f(x) = x^2, let's assume f(x1) = f(x2):
x1^2 = x2^2
Taking the square root of both sides, we get:
|x1| = |x2|
Now, we can see that for any positive value of x, there are two different inputs (x and -x) that give the same output. Therefore, the function is not injective.
Surjectivity:
To check if a function is surjective, we need to ensure that every element in the codomain has a pre-image in the domain. In this case, the function f(x) = x^2 maps every real number x to a non-negative real number.
However, there are non-negative real numbers that do not have a real square root. For example, there is no real number x such that x^2 = -1. Therefore, the function is not surjective.
Since the function is neither injective nor surjective, it does not have an inverse. Hence, the correct answer is option C: does not exist since it is not a bijection.
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Community Answer
If f is a function defined from R to R, is given by f(x) = x2then f-1(...
It is not a one one function hence Inverse does not exist.
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If f is a function defined from R to R, is given by f(x) = x2then f-1(x) is given by?a)1/(3x-5)b)(x+5)/3c)does not exist since it is not a bijectiond)none of the mentionedCorrect answer is option 'C'. Can you explain this answer?
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