Question

Prove that on \(R\), the function \(f(x)=e^{2x}\) is increasing.

Hint:

Show f'(x)=2e^{2x} is always positive on R.

Answer 1

Step 1: Key Formula or Approach:
A differentiable function \(f\) is increasing on an interval if \(f'(x) > 0\) for every \(x\) in that interval.

Step 2: Differentiating \(f(x)\):
\[ f'(x) = \frac{d}{dx}e^{2x} = 2e^{2x} \]

Step 3: Sign of \(f'(x)\):
The exponential function \(e^{2x}\) is positive for every real \(x\) (an exponential is never zero or negative), and multiplying by 2 keeps it positive. So \(f'(x) = 2e^{2x} > 0\) for all \(x \in R\).

Final Answer:
Since \(f'(x) > 0\) for every \(x \in R\), \(f(x)=e^{2x}\) is strictly increasing on \(R\). \[ \boxed{f'(x)=2e^{2x}>0 \implies f \text{ is increasing on } R} \]