Question

Prove that the given function \(f(x)=3x+17\) is increasing on \(\mathbb{R}\).

Hint:

Show \(f'(x)>0\) everywhere, or directly compare \(f(x_2)-f(x_1)\) for \(x_2>x_1\).

Answer 1

Step 1: Understanding the Concept:
A differentiable function is (strictly) increasing on an interval if its derivative is positive throughout that interval.

Step 2: Differentiating:
\(f'(x)=\dfrac{d}{dx}(3x+17)=3\).

Step 3: Checking the sign:
\(f'(x)=3>0\) for every real \(x\) — the derivative doesn't even depend on \(x\), so it's positive everywhere without exception.

Final Answer:
Since \(f'(x)>0\ \forall x\in\mathbb{R}\), \(\boxed{f(x)=3x+17\text{ is strictly increasing on }\mathbb{R}}\).