Q.Show that the function given by $f(x) = e^{2x}$ is increasing on $\mathbf{R}$.

The key idea is monotonicity of exponential: the exponential function $e^{x}$ is strictly increasing, and composing it with a positive linear transformation preserves that property.

Step 1: Compute the derivative of $f(x) = e^{2x}$:

Step 2: For any real $x$, $e^{2x} > 0$, so $2e^{2x} > 0$. Hence $f'(x) > 0$ for all $x \in \mathbf{R}$.

Step 3: A function with a positive derivative on an interval is strictly increasing on that interval. Since $f'(x) > 0$ everywhere on $\mathbf{R}$, $f$ is strictly increasing on $\mathbf{R}$.

The core idea here is monotonicity of exponential functions. An exponential function $a^x$ with $a > 1$ is always increasing — its graph rises as $x$ increases. Here, $e^{2x}$ is just a composition: the inner function $2x$ is linear and increasing, and the outer function $e^t$ is also increasing. The composition of two increasing functions is increasing. But the cleanest, most exam-ready method is to use the derivative test.

Let’s work through it step by step.

1. Recall the derivative test for monotonicity.

   A function $f$ is increasing on an interval if $f'(x) \geq 0$ for all $x$ in that interval, and strictly increasing if $f'(x) > 0$ for all $x$. For $\mathbf{R}$, we just need to check the sign of $f'(x)$ everywhere.

2. Differentiate $f(x) = e^{2x}$.

   Using the chain rule:

3. Analyze the sign of $f'(x)$. The exponential function $e^{2x}$ is always positive for any real $x$ — it never touches zero, never becomes negative. Multiplying by the positive constant $2$ keeps it positive. So:

4. Conclude monotonicity. Since the derivative is strictly positive everywhere, $f$ is strictly increasing on the entire real line.

Method: Proving Monotonicity of a Composite Exponential Function

When the function is an exponential raised to a linear (or otherwise differentiable) expression, such as $e^{g(x)}$, monotonicity is proved by differentiating with the chain rule and using the fact that an exponential expression is always positive, regardless of what its exponent evaluates to.

Steps

Step 1: Identify the composite structure

Recognise the function as $f(x) = e^{g(x)}$ (outer function $e^{t}$, inner function $g(x)$).

Step 2: Differentiate using the chain rule

Do not drop the inner derivative $g'(x)$ — this is the step most often missed.

Step 3: Argue the sign of each factor separately

The factor $e^{g(x)}$ is positive for every real value of $g(x)$ — this is a property of the exponential function itself, true no matter how large or small (even very negative) the exponent is. So the sign of $f'(x)$ is entirely decided by the sign of $g'(x)$.

Step 4 (Applying to this problem): Combine and conclude

If $g'(x)$ is itself a positive constant (as it is for a linear inner function), the product $f'(x) = e^{g(x)} \cdot g'(x)$ is positive for every real $x$, so by the Increasing Function Test, $f$ is strictly increasing on all of $\mathbf{R}$.

This pattern generalises: any $a^{g(x)}$ with base $a>1$ is increasing wherever $g'(x) > 0$, since $a^{g(x)}$ is always positive.

Common Mistakes

Mistake 1: Forgetting the chain-rule factor when differentiating $e^{2x}$

Why it's wrong: Writing $f'(x)=e^{2x}$ (dropping the factor of $2$) ignores the chain rule — the derivative of the inner function $2x$, which is $2$, must multiply the exponential. Correct approach: always write $\frac{d}{dx}e^{g(x)} = e^{g(x)}\cdot g'(x)$, so here $f'(x) = 2e^{2x}$.

Mistake 2: Thinking a fast-growing function might not be increasing for very negative $x$

Why it's wrong: Because $e^{2x}$ grows extremely fast for large positive $x$, students sometimes assume the function could flatten or decrease for very negative $x$. But $e^{2x}$ is positive for every real $x$, however negative — it just becomes a very small positive number, never zero or negative, so $f'(x)=2e^{2x}$ never fails to be positive. Correct approach: argue the sign of $e^{2x}$ abstractly (always positive, for any real exponent) rather than checking specific numerical values of $x$.