Secant slope on a rational function — same recipe, new function shape.
Even though $f(x) = 2/x$ looks different from the polynomials in Q1–Q3, the secant-slope recipe is identical:
The reason the recipe doesn't change: a secant slope is "rise over run" between two points on any curve. The function's shape doesn't matter — the formula is universal. What does require a different tool is going from a secant slope to a tangent slope (the derivative), which is why the difference quotient $\frac{f(x+h)-f(x)}{h}$ for $2/x$ needs the quotient rule later.
For this problem you don't need any rules — just arithmetic. The new thing to notice is the negative sign in $f(-1)$: $f(-1) = 2/(-1) = -2$, not $+2$.
For any function $f$ and two specific $x$-values $a$ and $b$, the slope of the secant line is:
$$m_{\text{sec}} \;=\; \frac{f(b) - f(a)}{b - a}$$(Three equivalent names: secant slope, average rate of change, difference quotient evaluated at $a, b$.)
Given $f(x) = \dfrac{2}{x}$, with $a = -1$ and $b = 2$.
Step 1 — Evaluate $f$ at each $x$-value.
$$f(-1) \;=\; \frac{2}{-1} \;=\; -2 \quad\quad \text{point: } (-1,\ -2)$$ $$f(2) \;=\; \frac{2}{2} \;=\; 1 \quad\quad \text{point: } (2,\ 1)$$Watch the sign: $2$ divided by $-1$ is $-2$, not $2$. Forgetting the negative here is the #1 mistake on this kind of problem.
Step 2 — Plug into the secant formula.
$$m_{\text{sec}} \;=\; \frac{f(2) - f(-1)}{2 - (-1)} \;=\; \frac{1 - (-2)}{2 + 1} \;=\; \frac{1 + 2}{3} \;=\; \frac{3}{3}$$ $$m_{\text{sec}} \;=\; 1$$Final answer:
Verification — does this slope make sense?
From $(-1, -2)$ to $(2, 1)$: rise $= 1 - (-2) = +3$, run $= 2 - (-1) = +3$. Rise over run = $3/3 = 1$. ✓
Slope 1 means the line goes up at a 45° angle. If you actually plotted the secant through $(-1,-2)$ and $(2,1)$, it would indeed be a "rise 1, run 1" line.
The secant slope recipe is the same no matter what $f$ looks like — the function's form only matters for the derivative. Here's a quick comparison:
| Secant slope | $m = \dfrac{f(b) - f(a)}{b - a}$ | works for any $f$ |
| Difference quotient | $\dfrac{f(x+h) - f(x)}{h}$ | works for any $f$ |
| Derivative | $\lim_{h \to 0}$ of the above | needs rules when $f$ is not a polynomial |
So: secant and average-rate-of-change problems stay arithmetic even when the function gets weird. Only the limit step is the hard part — and that's where the quotient rule, product rule, etc. come in later.
Find the slope of the secant line to $f(x) = \dfrac{3}{x}$ between $x = 1$ and $x = 3$.
Hint: $f(1) = ?$, $f(3) = ?$. Both are positive here, so no sign traps. Just plug, subtract, divide.
$f(1) = 3/1 = 3$, $f(3) = 3/3 = 1$.
$m_{\text{sec}} = \dfrac{f(3) - f(1)}{3 - 1} = \dfrac{1 - 3}{2} = \dfrac{-2}{2} = -1$
Sanity check: from $(1, 3)$ to $(3, 1)$ — going right by 2 and down by 2, so the slope is negative. Magnitude is 1. ✓
Send Q5 when ready. The recipe is locked in — same two steps regardless of $f$. The harder question comes when you need the derivative (the limit), not just the secant slope — that's when quotient rule enters the chat.