# Single Variable Calculus, Chapter 1, 1.1, Section 1.1, Problem 23

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Evaluate the difference quotient $\displaystyle \frac{f (3+h)-f(3)}{h}$ for the function $f(x) = 4 + 3x - x^2$
\begin{aligned} \displaystyle \frac{f (3+h)-f(3)}{h} &= \frac{4+3(3+h)-(3+h)^2 - [4+3(3)-(3)^2]}{h} && ( \text{Substitute f(3+h) and f(3) to the function f(x), then divide it by h} ) \\ \\ &= \frac{4+9+3h-[9+6h+h^2]-4-9+9}{h} &&( \text{ Simplify the equation})\\ \\ &= \frac{3h-6h-h^2}{h} &&( \text{ Combine like terms})\\ \\ &= \frac{-3h-h^2}{h} = \frac{\cancel{h}(-3 - h)}{\cancel{h}} &&( \text{ Factor the numerator and cancel out like terms})\\ \\ & = -3 -h \end{aligned}