Let f(x)= x-3/x, x>0. Show f is one to one and find the inverse for f.

Question

Calculus 2 homework question again. Need help showing that it is one to one. Would appreciate showing the inverse but I think I can handle that. Thank you for an help.

Answer 1

f'(x) = 1 + 3/x^2 > 0 on (0, ∞), so f is 1-1

Note that I am assuming that this is not (x-3)/x, but x - 3/x

The range will be (-∞, ∞). As x goes to 0 from the positive direction, y goes to -∞. As x goes to ∞, y goes to ∞

To get, the inverse, you solve for x in terms of y.

y = x - 3/x

Then, y = (x^2 - 3)/x, or x^2 - 3 = xy, or x^2 - xy - 3 = 0, or x = y/2 ± 1/2√(y^2 + 12)

As f(x) is defined for x on (0, ∞), we take the positive root. x = y/2 + 1/2√(y^2 + 12)

Then, we just switch x and y to write the inverse function

f^-1(x) = x/2 + 1/2√(x^2 + 12) The domain of the inverse function will be (-∞, ∞), and the range will be (0, ∞)