f: [0, ∞) --> [0, ∞)
f(x) = x/(1+x)
i know this will be one-one since f'(x) is strictly positive and as for onto, someone told me that if i prove f(x) = y then this will be onto so i gave it a try:
from f(x) i got x = y/(1-y)
then i substituted this value of x in f(x) which indeed gave me f(x) = y but the final answer says the function is not onto. so either the answer is wrong(which i doubt) or i am doing it wrong.
So you have that $f(x) = \frac{x}{1+x}$. You can always take the derivative, but we will show injectivity in a different manner.
Let $f(x)=f(y)$, then $\frac{x}{1+x} = \frac{y}{1+y}$. Cross multiply, you get $x + xy=y+xy$, from which cancelling $xy$, you get $x=y$. There is no need to even think about derivatives.
On the other hand, $f$ is not onto. To see this, note that $f(x) = \frac{x}{1+x} = 1-\frac{1}{1+x}$. If $0 \leq x < \infty$, then $1 \leq 1+x<\infty$, so that $1 \geq \frac{1}{1+x}>0$ , so that $-1 \leq \frac{-1}{1+x} < 0$, so that $0 \leq 1-\frac 1{1+x} < 1$, so the range of the function is in fact $[0,1)$, and not $[0,\infty)$ as original codomain.
If you want that $f(x)=y$, you get that necessarily $$x=\frac{y}{1-y}$$ So for all $y\neq1$ there exists $x$ such that $f(x)=y$. What if $y=1$? You want to find $x$ such that $$\frac{x}{x+1}=1$$ So $$x=x+1$$ Which doesn't hold for any $x\in\Bbb R$.
Note that $$ f(x)=\frac{x}{x+1}=1-\frac{1}{x+1}<1 $$ for all $x\geq0$.
