Examples of Continuous Strictly Increasing Functions on ℝ

The title asks for examples of continuous strictly increasing functions on $\mathbb{R}$, but does not state any additional condition. A standard example is $g(x)=\arctan x$. If the intended extra condition is that the function be not onto (equivalently, bounded on at least one side), this is a suitable example.

The checks are straightforward: arctangent is continuous and has positive derivative everywhere, so it is strictly increasing. Its values lie between $-\pi/2$ and $\pi/2$, so its range is not all of $\mathbb{R}$. The details below also show how to distinguish this example from other common possibilities.

A bounded, strictly increasing example

Define

$$g(x)=\arctan x, \qquad x\in\mathbb{R}.$$

The derivative is

$$g'(x)=\frac{1}{1+x^2}>0$$

for every real $x$. By the mean value theorem, a differentiable function with positive derivative on an interval is strictly increasing there. Thus, whenever $a<b$, we have $g(a)<g(b)$.

At the same time,

$$-\frac{\pi}{2}<\arctan x<\frac{\pi}{2}$$

for every real $x$. In fact, its range is exactly $(-\pi/2,\pi/2)$. Therefore $g$ is bounded and is not onto $\mathbb{R}$, even though it is continuous and strictly increasing.

For a concrete comparison, $g(0)=0$ and $g(1)=\pi/4$, so the output increases as the input increases. But no input gives an output of $2$, since $2>\pi/2$.

If the missing condition was different

The end of the title matters: continuity and strict increase alone do not determine whether a function is onto, bounded, or differentiable. Here are examples for several likely conditions.

Possible extra condition	Example	Why it works
Not onto $\mathbb{R}$; bounded	$g(x)=\arctan x$	Its range is $(-\pi/2,\pi/2)$.
Onto $\mathbb{R}$	$g(x)=x^3$	It is continuous, strictly increasing, and has range $\mathbb{R}$.
Not differentiable everywhere	$g(x)=x+	x

For the last example, write it piecewise:

$$g(x)=\begin{cases} x/2,&x<0,\\ 3x/2,&x\geq 0. \end{cases}$$

Both pieces have positive slope, and values on the negative side are below the values on the nonnegative side. This proves strict increase across the join as well as within each piece. The left and right derivatives at zero are $1/2$ and $3/2$, respectively, so the derivative at zero does not exist. The function is nevertheless continuous there, since both pieces approach zero.

A useful test for being onto

For any continuous strictly increasing function on $\mathbb{R}$, its limits at the two ends of the domain control its range. If

$$\lim_{x\to-\infty}g(x)=-\infty \quad\text{and}\quad \lim_{x\to\infty}g(x)=\infty,$$

then the intermediate value theorem implies that $g$ takes every real value. Strict increase also ensures that each value is taken at most once, so in this case $g$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$.

For $g(x)=\arctan x$, the corresponding limits are $-\pi/2$ and $\pi/2$, not $-\infty$ and $\infty$. That is why it is not onto. By contrast, $x^3$ tends to $-\infty$ and $\infty$ at the respective ends and is onto.

Common mistakes

• Assuming strict increase implies onto. It does not: arctangent is strictly increasing but has a bounded range.
• Using a constant or flat interval to make a function bounded. A constant function is not strictly increasing, and neither is a function with a flat interval.
• Confusing “not differentiable everywhere” with “nowhere differentiable.” The piecewise-linear example above fails to be differentiable at one point only. A condition demanding nowhere differentiable needs a different, more specialized construction.
• Ignoring the missing condition. If the intended ending was something other than “not onto” or “bounded,” check that the proposed example satisfies that specific property too.