A negative reciprocal — two sign traps in one problem.
The recipe is the same as Q2 (same function type, same "find and simplify" wording), but the negative sign on top creates two new traps:
The skill being tested: you have to do the difference of two negative fractions carefully. Every step looks like a place to drop a sign.
The general difference quotient:
$$\frac{f(x+h) - f(x)}{h}$$For any $f$ — including rational functions like $-1/x$. The function's shape doesn't change the recipe; it just changes how messy steps 2 and 3 get.
You'll also need the recipe for subtracting fractions with different denominators:
$$\frac{a}{b} - \frac{c}{d} \;=\; \frac{a \cdot d - c \cdot b}{b \cdot d}$$That "common denominator" trick is the heart of step 3 for any rational function.
Given $f(x) = \dfrac{-1}{x}$.
Step 1 — Write the difference quotient.
$$\frac{f(x+h) - f(x)}{h}$$Step 2 — Compute $f(x+h)$.
Substitute $(x+h)$ for $x$ in the formula $f(x) = -1/x$:
$$f(x+h) \;=\; \frac{-1}{x+h}$$Step 3 — Subtract $f(x)$ — this is where the sign work happens.
Subtract the two fractions, getting a common denominator first. The second fraction gets distributed through the minus sign, flipping its sign:
$$f(x+h) - f(x) \;=\; \frac{-1}{x+h} - \frac{-1}{x} \;=\; \frac{-1}{x+h} + \frac{1}{x}$$Now the two fractions have different denominators ($x+h$ and $x$). Use the common-denominator recipe $\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad + cb}{bd}$:
$$= \frac{(-1) \cdot x \;+\; 1 \cdot (x+h)}{(x+h) \cdot x} \;=\; \frac{-x + x + h}{x(x+h)} \;=\; \frac{h}{x(x+h)}$$Step 4 — Divide by $h$ and simplify.
$$\frac{1}{h} \cdot \frac{h}{x(x+h)} \;=\; \frac{h}{h \cdot x(x+h)} \;=\; \frac{1}{x(x+h)}$$Final answer:
Each wrong answer corresponds to a specific place where the algebra went off the rails. Read them carefully — this is the half of the quiz that teaches the most.
For this kind of rational-function problem, the friction is in step 3. Here's the recipe to internalize:
For $f(x) = 1/x$ (positive version), the same recipe gives $\dfrac{1}{x+h} - \dfrac{1}{x} = \dfrac{x - (x+h)}{x(x+h)} = \dfrac{-h}{x(x+h)}$, and after dividing by $h$: $\dfrac{-1}{x(x+h)}$. So the answer for $f(x) = 1/x$ is the negative of the answer for $f(x) = -1/x$. Sanity check: it should be — they're the same function with one sign flipped, so the difference quotient should also be one sign flipped.
Find and simplify the general difference quotient of $g(x) = \dfrac{1}{x+1}$.
Hint: $g(x+h) = \dfrac{1}{x+h+1} = \dfrac{1}{(x+1)+h}$. The denominators will be $(x+1)+h$ and $(x+1)$ — call them $u + h$ and $u$ where $u = x+1$, and the algebra is the same as Q5.
Step 2: $g(x+h) = \dfrac{1}{x+h+1}$
Step 3: $g(x+h) - g(x) = \dfrac{1}{x+h+1} - \dfrac{1}{x+1} = \dfrac{(x+1) - (x+h+1)}{(x+h+1)(x+1)} = \dfrac{x + 1 - x - h - 1}{(x+h+1)(x+1)} = \dfrac{-h}{(x+h+1)(x+1)}$
Step 4: $\dfrac{1}{h} \cdot \dfrac{-h}{(x+h+1)(x+1)} = \dfrac{-1}{(x+h+1)(x+1)}$
Sanity check: $g$ is the Q5 function shifted left by 1, so the answer should look like the Q5 answer with $x$ replaced by $x+1$. Q5's answer was $\dfrac{1}{x(x+h)}$; this is $\dfrac{-1}{(x+1)(x+1+h)}$. The shift worked; the extra negative sign is because $g$ is the positive version, not the negative version of Q5. ✓
Send Q6 if you have it — same recipe, harder $f$. For any rational function of the form $\dfrac{c}{x}$ (where $c$ is a constant), the difference quotient always has the structure $\dfrac{\text{something}}{x(x+h)}$ — it's just the "something" that changes with $c$ and the sign. Once you can do this for $c = -1$, $c = 1$, and $c = 3$, you can do it for any $c$.