Is this function injective, surjective, or both?

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The function f(x) = (x+1)/(x+2) for x ≠ -2 and f(x) = 1 for x = -2 is analyzed for injectivity and surjectivity. The derivative f'(x) is positive for all x, indicating that the function is strictly increasing. The limits as x approaches both infinities converge to 1, and there is a horizontal asymptote at x = -2, leading to a range that includes all real numbers. The function is determined to be bijective, meaning it is both injective and surjective. The original poster considers sharing their reasoning for the benefit of others.
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Homework Statement



The following function f is a function from R to R. Determine whether f is injective (one-to-one), surjective (onto), or both. Please give reasons.

Homework Equations



f(x) = (x+1)/(x+2) if x != -2
f(x) = 1 when x = 2

The Attempt at a Solution



f'(x) = 1/(x+2)2 > 0 for all x

and the limits at both infinities are 1 using l'hopital.

So the way I see it is the function grows from 1+ when x is a large negative, and then there's a horizontal assymptote at x = -2 so just before x = -2 f(x) tends to infinity and just after x = -2 the f(x) goes from negative infinity and gradually increases to 1. and then ofcourse at the point x = - 2 f(x) = 1 because that value has been forced in the definition.

So now by visualising the graph I have a strong suspicion that this function is bijective, but I have no idea how to prove it 'analytically'
 
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OK nevermind i figured it out. Is there a way to remove my original post? Or should I post my reasoning incase it could be helpful to others?
 
Last edited:
Usually the latter is preferred
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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