Find and simplify the difference quotient (f(x + h) - f(x))/h, where h is not 0, for f(x) = 8 - 11x.

Step-by-step solution

1. 1

   Build $f(x+h)$ by substitution, using brackets. Replace every $x$ in the rule by the whole quantity $x+h$:

   $f(x+h) = 8 - 11(x+h) = 8 - 11x - 11h$

   Writing $8 - 11x + h$ — distributing the $-11$ to only the first term — is the most common slip.

2. 2

   Form the numerator and watch the subtraction. Subtracting $f(x)$ means subtracting the entire expression, so both its terms flip sign:

   $f(x+h) - f(x) = (8 - 11x - 11h) - (8 - 11x) = 8 - 11x - 11h - 8 + 11x$

3. 3

   Cancel everything without an $h$. The $8$ and $-8$ cancel, and $-11x$ and $+11x$ cancel:

   $f(x+h) - f(x) = -11h$

   That every $x$ disappears is a signal the function is linear.

4. 4

   Divide by $h$. Since $h \neq 0$, the division is legal:

   $\frac{-11h}{h} = -11$

   No $h$ and no $x$ survive.

5. 5

   Interpret the constant result. The difference quotient is the slope of the secant line through $(x, f(x))$ and $(x+h, f(x+h))$. For a straight line every secant is the line, so the quotient must be the constant slope $-11$ — matching the coefficient of $x$ in $f(x) = 8 - 11x$. Taking $h \to 0$ then gives $f'(x) = -11$, as expected.

6. 6

   Contrast with a quadratic. For $f(x) = 2x^2 - 3x + 4$ the same procedure gives $\dfrac{f(x+h)-f(x)}{h} = 4x + 2h - 3$, which still contains both $x$ and $h$. The presence of a leftover $h$ is normal for non-linear functions; its absence here is exactly what identifies $8-11x$ as linear.

7. 7

   Verify numerically. Take $x = 2$, $h = 0.1$: $f(2.1) = 8 - 23.1 = -15.1$ and $f(2) = 8 - 22 = -14$, so the quotient is $\dfrac{-15.1 + 14}{0.1} = \dfrac{-1.1}{0.1} = -11$ ✓, independent of $h$ as claimed.