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: Working from the definition directly:
Take any \(x_1<x_2\) in \(\mathbb{R}\).

Step 2: Comparing function values:
\(f(x_2)-f(x_1)=(3x_2+17)-(3x_1+17)=3(x_2-x_1)\). Since \(x_2-x_1>0\), this difference is positive, i.e. \(f(x_2)>f(x_1)\).

Final Answer:
So \(x_1<x_2\Rightarrow f(x_1)<f(x_2)\), which is exactly the definition of \(\boxed{\text{strictly increasing}}\).