Average rate of change via the secant line slope

Substitute into the difference quotient: $\frac{f(x+h)-f(x)}{h} = \frac{[3(x+h)-5]-[3x-5]}{h}.$ Expanding the numerator gives $3x + 3h - 5 - 3x + 5 = 3h$.

So the quotient simplifies to $\frac{3h}{h} = 3$ for any $h \neq 0$. This shows the average rate of change is always $3$, regardless of $x$ or $h$.

From part (a), the difference quotient simplifies to $3$ for all $h \neq 0$, so $f'(x) = \lim_{h\to 0} 3 = 3.$

This equals the average rate of change computed in part (a) exactly, for every $x$ and every $h$. This confirms that for a linear function, the average rate of change over any interval equals the instantaneous rate of change everywhere — both are the constant slope $3$.