Question

Let f:R→R be the function defined by f(x)= x/x²+1 for any x∈R.

Fill in the blanks in the blocks below, all labelled by capital-letter Roman numerals, with appropriate words so that they give respectively a proof for the statement (C) and a proof for the statement (D). (The 'underline' for each blank bears no definite relation with the length of the unswer for that blank.) (a) Here we prove the statement (C) : (C) The function f is not surjective.
[We want to verify that f is not surjective. This amounts to verifying the statement (I) _____y₀ ∈R such that (II) _________x∈R such that (III) _________.]
(IV) _________y₀=1
We verify, using the method of proof-by-contradiction, that (V)________ f(x)≠y₀:

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(VI)-_____________ it were true that (VII)_________
Then x₀/x₀²+1 =f(x₀)=y₀​=1.
Therefore (VIII)_________ =x ₀²-x₀+1= (IX) >0.
Contradiction arises. It follows that (X)__________
(b) Here we prove the statement (D) :
(D) The function f is not injective.
[We want to verify that f is not injective. This amounts to verifying the statement (I)_____ x₀ ,ω₀ ∈R such that f(x₀) (II)_____ and (III)____ ]
Take x₀ = 1/2, (IV)_________ . Note that x₀,ω₀ ∈R. Also note that (V) We have f(x₀)= (1/2)/ (1/2)²+1=2/5 and (VI) __________Then f(x₀)= (VII)__________ It follows that (VIII)______

Answer 1

To prove that the function f(x) = x/(x²+1) is not surjective, we choose y₀ = 1 and show that there is no x in the domain that maps to y₀. To prove that f is not injective, we choose x₀ = 1/2 and y₀ = 2/5 and show that there exists another x in the domain that maps to y₀.

Step-by-step explanation:

In order to prove that the function f(x) = x/(x²+1) is not surjective, we need to find a value y₀ in the range of the function such that there is no x in the domain that maps to y₀. Let's go through the steps:

1. (I) Choose y₀ = 1
2. (II) Assume there exists an x in the domain such that f(x) = y₀
3. (III) Solve the equation x/(x²+1) = 1 for x
4. (IV) Simplify the equation x² - x + 1 > 0
5. (V) Use the method of proof-by-contradiction to assume that x/(x²+1) = 1
6. (VI) If x/(x²+1) = 1, then x² - x + 1 = 0
7. (VII) Solve the equation x² - x + 1 = 0 for x and find the solutions x₀ that make the equation true
8. (VIII) Simplify the equation and find that x₀² - x₀ + 1 > 0, which is a contradiction
9. (IX) Since the assumption led to a contradiction, we conclude that there is no x in the domain such that f(x) = y₀
10. (X) Therefore, the function f is not surjective

To prove that the function f is not injective, the steps are as follows:

1. (I) Choose x₀ = 1/2 and y₀ = 2/5
2. (II) Assume there exists another x in the domain such that f(x) = f(x₀) = y₀
3. (III) Simplify the equation (x)/(x²+1) = (1/2)/(1/2)²+1 = 2/5
4. (IV) Notice that x₀ also satisfies the equation x/(x²+1) = 2/5
5. (V) Thus, we have f(x₀) = f(x) = 2/5
6. (VI) Since there exists another x that maps to the same value y₀, the function f is not injective
7. (VII) Therefore, the function f is not injective