- #1

bigplanet401

- 104

- 0

## Homework Statement

Show that

[tex]

f(x) = \sqrt{5x+2}

[/tex]

is one-to-one.

## Homework Equations

If f' >0 or f' < 0 everywhere on f's domain, f is one-to-one.

## The Attempt at a Solution

[tex]

f^\prime = \frac{5}{2} \frac{1}{\sqrt{5x +2}} = \frac{5}{2f}

[/tex]

f' is positive on (-2/5, infinity) but is undefined at x = -2/5. I can therefore say that f is one-to-one on this interval, but what about the point x=-2/5? The derivative is undefined there (the left-hand limit does not exist). There might be a way to argue that f is increasing on [-2/5, infinity), but I don't know how to do so since I really don't know if f(-2/5) = 0 is the minimum value.