Show that f(x) = sqrt(x) / (sqrt(x) + 1) with D = [0, infinity) is one-to-one?

Answer 1

Let h(x)=sqrt(x)
Let g(x)=x/(x+1)

Then f(x) = g(h(x))

h(x) is 1-1 on the domain since h(x)=h(y) means sqrt(x)=sqrt(y). Squaring both sides gives x=y.
So h is 1-1.

Now if g(x) = g(y) then, using an equivalent g(x):

g(x) = 1 - [1/(x+1)]

1-[1/(x+1)] = 1-[1/(y+1)]

From this, we see that 1/(x+1) = 1/(y+1).

Inverting, we see that x+1 = y+1, or x=y.

So g and h are both 1-1, so g(h(x))=f(x) is 1-1.

Answer 2

We have to show f is either strictly increasing or strictly decreasing in [0, oo). Let g(x) = sqrt(x). Then, g is strictly increasing in [0, infinity). If you divide the numerator and the denominator by g(x), then th function f can be written for x >0 as

f(x) = 1/(1 +1/g(x))

Since g is strictly increasing and positive, so is 1/g. Hence, , the same is true of 1 + 1/g. And since f = 1/(1 +1/g), it follows f is strictly increasing in (0, oo). Therefore f is one-to-one in (0, oo). And since f is positive in (0, oo) and f(0) = 0, it follows f is strictly increasing - and, therefore, one -to-one - in [0, oo)

Note that this is true whenever g is non-negative and strictly increasing, the conclusion is not restricted to the case g(x) = sqrt(x).

Answer 3

sorry only made it as high as algebra one and i took it again in summer school lol