Question

Show that $f : \mathbb{R} \to \mathbb{R}$ defined as $f(x) = \frac{x}{\sqrt{1+x^2}}$ is one-one but not onto.}

Hint:

To show a function is not onto, it's enough to find its range. If the range is a bounded subset of an infinite codomain, it's definitely not onto. For one-one, if the derivative $f'(x)$ is strictly positive or strictly negative throughout the domain, the function is injective.

Answer 1

Concept:
• A function is one-one (injective) if $f(x_1) = f(x_2) \implies x_1 = x_2$.
• A function is onto (surjective) if for every $y$ in the codomain, there exists an $x$ in the domain such that $f(x) = y$.

Step 1: Test for One-One property
Let $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$: \[ \frac{x_1}{\sqrt{1+x_1^2}} = \frac{x_2}{\sqrt{1+x_2^2}} \] Squaring both sides: \[ \frac{x_1^2}{1+x_1^2} = \frac{x_2^2}{1+x_2^2} \] \[ x_1^2(1+x_2^2) = x_2^2(1+x_1^2) \] \[ x_1^2 + x_1^2x_2^2 = x_2^2 + x_2^2x_1^2 \] \[ x_1^2 = x_2^2 \implies |x_1| = |x_2| \] Since $f(x)$ has the same sign as $x$, for the equality to hold, $x_1$ and $x_2$ must have the same sign.
Thus, $x_1 = x_2$. The function is one-one.

Step 2: Test for Onto property
Let $y = \frac{x}{\sqrt{1+x^2}}$.
Notice that for any $x \in \mathbb{R}$, $x^2 < 1 + x^2$, which implies $\sqrt{x^2} < \sqrt{1 + x^2}$, so $|x| < \sqrt{1+x^2}$.
This means $|f(x)| < 1$ for all $x$.
The range of the function is the interval $(-1, 1)$.

Step 3: Conclude based on codomain
The codomain is given as $\mathbb{R}$. Since the range $(-1, 1) \neq \mathbb{R}$, there are values in the codomain (like $y=2$) that have no corresponding $x$. Thus, the function is not onto.