Q.Find the maximum and the minimum values, if any, of the function given by $f(x) = x,\ x \in (0, 1)$.

Idea: On an open interval a strictly increasing function reaches neither a largest nor a smallest value, because the endpoints are excluded.

$f(x)=x$ on $(0,1)$ is strictly increasing.

• No minimum: for any $c\in(0,1)$, the smaller number $\tfrac{c}{2}$ is still in $(0,1)$ and $f\!\left(\tfrac{c}{2}\right)<f(c)$. So no value is the least.
• No maximum: for any $c\in(0,1)$, the larger number $\tfrac{c+1}{2}$ is still in $(0,1)$ and $f\!\left(\tfrac{c+1}{2}\right)>f(c)$. So no value is the greatest.

The function gets arbitrarily close to $0$ and $1$ but never attains them.

The key idea

A maximum value is a value the function actually attains that is at least as large as every other value; a minimum is one it attains that is at least as small. On a closed interval $[a,b]$ a continuous function is guaranteed both. On an open interval that guarantee disappears, and this example shows exactly how.

Understand the domain

The interval $(0,1)$ means $0<x<1$; the endpoints $0$ and $1$ are excluded. So $f(0)$ and $f(1)$ are simply not part of the function.

Why there is no minimum

Suppose someone claims the minimum is attained at some $c\in(0,1)$, i.e. $f(c)=c$ is the smallest value. But $\dfrac{c}{2}$ is also in $(0,1)$ (it is positive and less than $c<1$), and

We found a smaller value, contradicting the claim. So no minimum exists.

Why there is no maximum

Similarly, suppose $f(c)=c$ were the largest value for some $c\in(0,1)$. Then $\dfrac{c+1}{2}$ lies strictly between $c$ and $1$, so it is in $(0,1)$, and

We found a larger value. So no maximum exists.

The bounds are approached, not reached

The values of $f$ fill the open interval $(0,1)$: they come arbitrarily close to $0$ and to $1$ but never equal them. (In higher language, $0$ and $1$ are the infimum and supremum, but they are not attained.) Had the domain been the closed interval $[0,1]$, then $f(0)=0$ would be the minimum and $f(1)=1$ the maximum — the whole difference is the inclusion of the endpoints.

Method: Deciding Whether a Function Attains a Maximum or Minimum

Before hunting for the largest or smallest value of a function, check whether the domain and the function's behaviour even guarantee one exists — many students jump straight to "compare endpoint values" without checking this first.

Steps

Step 1: Note the type of interval

A continuous function on a closed and bounded interval $[a,b]$ is guaranteed to attain both a maximum and a minimum. On an open interval $(a,b)$, or an unbounded one, this guarantee disappears — the endpoints are excluded, so a value the function only approaches near an endpoint is never actually reached.

Step 2: Check monotonicity

If $f'(x)$ keeps one sign throughout the interval (always positive or always negative), $f$ is strictly monotonic there. A strictly monotonic function's largest and smallest values, if they exist at all, can only occur at the endpoints — so on an open interval where the endpoints are excluded, neither is ever attained.

Step 3: Prove non-attainment by producing a "better" nearby point

For any candidate value $f(c)$ with $c$ inside an open interval, show that a nearby point still inside the domain gives a strictly larger (or smaller) value. If $f$ is increasing, the point $\frac{c+b}{2}$ (still inside $(a,b)$) beats $f(c)$ on the upper side, and $\frac{a+c}{2}$ beats it on the lower side. Since this works for every candidate $c$, no maximum or minimum can exist.

Step 4: Applying to this problem

$f(x)=x$ is strictly increasing on every interval. On $(0,1)$ the endpoints $0$ and $1$ are excluded, so by Steps 2–3 neither the maximum nor the minimum is attained — the values merely get arbitrarily close to $0$ and $1$.

Common Mistakes

Mistake 1: Assuming a monotonic function must have both a maximum and a minimum

Why it's wrong: this only holds on a closed interval (a continuous function on $[a,b]$ is guaranteed to attain its extremes there). On the open interval $(0,1)$, the endpoints $0$ and $1$ are excluded, so a strictly increasing function like $f(x)=x$ never actually reaches its smallest or largest possible value. Correct approach: always check whether the interval is open or closed before concluding extrema exist.

Mistake 2: Answering "minimum $=0$, maximum $=1$"

Why it's wrong: this treats the interval as if it were the closed $[0,1]$, silently including the excluded endpoints. $f(0)$ and $f(1)$ are simply not defined for this function on $(0,1)$. Correct approach: state clearly that no maximum and no minimum exist, rather than reporting the excluded boundary values as if they were attained.