Q.Let $f$ be a function defined on $[a, b]$ such that $f'(x) > 0$, for all $x \in (a, b)$. Then prove that $f$ is an increasing function on $(a, b)$.

The key idea is the Mean Value Theorem (MVT). If a function is differentiable on $(a, b)$ and continuous on $[a, b]$, then for any two points $x_1 < x_2$ in $(a, b)$, there exists some $c \in (x_1, x_2)$ such that:

Since $f'(x) > 0$ for all $x \in (a, b)$, we have $f'(c) > 0$ for this particular $c$. Therefore:

$\frac{f(x_2) - f(x_1)}{x_2 - x_1} > 0$ …

The core idea here is deceptively simple: if the derivative is positive everywhere, the function must be rising. But why is that logically airtight? The derivative only tells us about instantaneous behaviour — what happens at a single point. To conclude something about the function over an entire interval, we need a bridge between local slope and global change. That bridge is the Mean Value Theorem.

The Mean Value Theorem says: if a function is continuous on $[p,q]$ and differentiable on $(p,q)$, then there is some point $c$ inside where the instantaneous slope equals the average slope over the whole interval. In symbols:

This is powerful because it ties the difference in function values directly to the derivative at some interior point.

Now, to prove $f$ is increasing on $(a,b)$, we need to show: whenever $x_1 < x_2$ (both in $(a,b)$), we have $f(x_1) < f(x_2)$. Let's walk through it.

1. Pick any two points $x_1$ and $x_2$ in $(a,b)$ with $x_1 < x_2$. Since $f$ is differentiable on $(a,b)$, it is also continuous on $[a,b]$ (differentiability implies continuity). So $f$ satisfies the conditions of the Mean Value Theorem on the closed interval $[x_1, x_2]$.

2. Apply the Mean Value Theorem to $f$ on $[x_1, x_2]$. There exists some $c$ in $(x_1, x_2)$ such that

3. Use the given condition: we know $f'(x) > 0$ for every $x$ in $(a,b)$. Since $c$ lies in $(x_1, x_2) \subset (a,b)$, it follows that $f'(c) > 0$.

4. Combine the facts: The denominator $x_2 - x_1$ is positive (because $x_2 > x_1$). So we have

Multiplying both sides by the positive number $x_2 - x_1$ gives

$f(x_2) - f(x_1) > 0 \quad \Longrightarrow \quad f(x_2) > f(x_1).$ …

Method: Using the Mean Value Theorem to Prove Monotonicity

This method teaches the reasoning pattern for turning a statement about the sign of the derivative at every point into a statement about how the function's values compare at two different points — exactly the gap the Mean Value Theorem (MVT) bridges.

Steps

Step 1: Recognise when MVT is the right tool

Whenever you're given information about $f'(x)$ everywhere on an interval but asked to conclude something about $f(x_1)$ versus $f(x_2)$ at two specific points, you need a theorem that connects the local (instantaneous slope) to the global (change in value) — that is exactly what MVT does. A derivative being positive at a point only says the function is rising right there; it takes MVT to extend that to "therefore $f(x_2) > f(x_1)$".

Step 2: State the hypotheses and verify them

MVT requires $f$ to be continuous on the closed interval and differentiable on the open interval. Here, $f$ differentiable on $(a,b)$ already guarantees continuity there, so for any $x_1 < x_2$ chosen inside $(a,b)$, $f$ is continuous on $[x_1,x_2]$ and differentiable on $(x_1,x_2)$ — the hypotheses hold.

Step 3: Apply MVT on the sub-interval

For the chosen $x_1 < x_2$, MVT guarantees some $c \in (x_1,x_2)$ with

Step 4: Combine the given sign information with the sign of the denominator …

Common Mistakes

Mistake 1: Assuming $f'(x)>0 \Rightarrow f$ increasing without proof

Many students treat this as "obvious" and simply assert it, without ever invoking MVT. Why it's wrong: the derivative's sign is only local information about the instantaneous slope at each point; nothing automatically extends that to a comparison of function values at two separate points — that logical bridge is precisely what the question asks you to build. Correct approach: explicitly apply MVT on $[x_1,x_2]$ to produce the link.

Mistake 2: Treating an endpoint as a possible value of $c$

A student may write $f'(a)$ or $f'(b)$ into the MVT conclusion, or apply the theorem directly on $[a,b]$ using $f'(a)$ or $f'(b)$. Why it's wrong: MVT only guarantees a point $c$ in the open interval between the two chosen points — endpoints are never valid candidates for $c$, and in this problem $f'$ isn't even given to exist at $a$ or $b$ themselves. Correct approach: apply MVT to an arbitrary interior sub-interval $(x_1,x_2)\subset(a,b)$, not to $[a,b]$ itself. …