Is f(x) = √(5x+2) one-to-one on (-2/5, infinity)?

[bigplanet401]

Homework Statement

Show that

<br /> f(x) = \sqrt{5x+2}<br />

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

<br /> f^\prime = \frac{5}{2} \frac{1}{\sqrt{5x +2}} = \frac{5}{2f}<br />

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.

 

[member 428835]

You have shown that $f$ is monotonically increasing on $(-2/5,\infty)$. This combined with the continuity of $f$ along $x \in [0,\infty)$ is sufficient to state that $f$ is one-to-one.

Also, you can look at the definition of one-to-one when dealing with $x=-2/5$ (since you used a theorem for all other points of existence), and recall that one-to-one functions never map distinct elements of its domain to the same element of its range.

 

[bigplanet401]

You have shown that $f$ is monotonically increasing on $(-2/5,\infty)$. This combined with the continuity of $f$ along $x \in [0,\infty)$ is sufficient to state that $f$ is one-to-one.

Also, you can look at the definition of one-to-one when dealing with $x=-2/5$ (since you used a theorem for all other points of existence), and recall that one-to-one functions never map distinct elements of its domain to the same element of its range.

Thanks! But without doing anything else, I'm not sure I can assume f(x) will never be zero except at x=-2/5.

But after thinking a bit more, I was able to use the fact that f(a) = f(-2/5) is defined and the mean value theorem to write

f(x) - f(a) = f'(c)(x - a) where c is in (a, x). Since f'(c) > 0 when x > a, the right-hand side will be positive and so f(x) > f(a) everywhere in [a, x) (with x > a).

 

[member 428835]

Thanks! But without doing anything else, I'm not sure I can assume f(x) will never be zero except at x=-2/5.

But after thinking a bit more, I was able to use the fact that f(a) = f(-2/5) is defined and the mean value theorem to write

f(x) - f(a) = f'(c)(x - a) where c is in (a, x). Since f'(c) > 0 when x > a, the right-hand side will be positive and so f(x) > f(a) everywhere in [a, x) (with x > a).

Very nice. You are in analysis, correct, hence the burden of proof?

 

[bigplanet401]

Very nice. You are in analysis, correct, hence the burden of proof?

This was a problem in my calculus textbook. There's a theorem that says increasing/decreasing functions have inverses, but I wasn't sure how to show that the function satisfied the hypothesis.

 

[member 428835]

I see. I suppose another way to solve could have been to simply set $f=0$ and solve for $x$. You'd then know what values of $x$ give the desired output. It seems then your theorem covers all other points.

 

[HallsofIvy]

Directly from the definition of "one-to-one": If f(x)= f(y) then \sqrt{5x+ 2}= \sqrt{5y+ 2}. Show that x= y.