The function isn't injective because $f(x)=f\left(\frac{1}{x}\right)$. for $x\neq 1$.

Obel1x wrote: $\star$ It is given the function $f: \left(\mathbb{R} - \{0\}\right) \to \mathbb{R}$ such that $f(x)=x+\frac{1}{x}$. Is this function injective ? Justify your answer. Goutham I guess you meant x to be different from 0. And can you please post the other problems as I am interested in the probs?

myro111 wrote: And can you please post the other problems as I am interested in the probs? http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=396494, he already did

@myro111: x should be unequal to 0 as well as 1. If $x = 1$, $f(1) = f(1)$ doesn't tell us anything about the injectivity of f.

Pascal96 wrote: @myro111: x should be unequal to 0 as well as 1. If $x = 1$, $f(1) = f(1)$ doesn't tell us anything about the injectivity of f. True,true!

$x$ cannot equal $0$ (undefined), $1$ ($f(1) = f(1)$ is trivial), or $-1$ ($f(-1) = f(-1)$ is also trivial).

@fractals: you're right . It's always good to be careful about these small things, because the people correcting these papers jump at every opportunity to cut points!

Notice that $f(x)=x+ \frac{1}{x}$ $f(\frac{1}{x})= \frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} +x$ $\Rightarrow f(x)=f(\frac{1}{x})$ Therefore the function is not injective $\blacksquare$