Function Difference Quotient 48Cc12

1. **State the problem:** We have the function $$f(x) = x^2 - x - 1$$ and need to find: (a) $f(a+h)$ (b) $\frac{f(a+h) - f(a)}{h}$ for $h \neq 0$ (c) The instantaneous rate of change of $f$ at $a=4$ 2. **Find $f(a+h)$:** Substitute $x = a+h$ into $f(x)$: $$f(a+h) = (a+h)^2 - (a+h) - 1$$ Expand the square: $$= a^2 + 2ah + h^2 - a - h - 1$$ This is the simplified form (no factoring): $$\boxed{a^2 + 2ah + h^2 - a - h - 1}$$ 3. **Find $\frac{f(a+h) - f(a)}{h}$:** First, find $f(a)$: $$f(a) = a^2 - a - 1$$ Compute the numerator: $$f(a+h) - f(a) = (a^2 + 2ah + h^2 - a - h - 1) - (a^2 - a - 1)$$ Simplify inside the numerator: $$= a^2 + 2ah + h^2 - a - h - 1 - a^2 + a + 1$$ $$= 2ah + h^2 - h$$ Now divide by $h$ (with $h \neq 0$): $$\frac{2ah + h^2 - h}{h}$$ Factor $h$ out of numerator terms: $$= \frac{h(2a + h - 1)}{h}$$ Cancel $h$: $$= \cancel{\frac{h}{h}}(2a + h - 1) = 2a + h - 1$$ So the simplified expression is: $$\boxed{2a + h - 1}$$ 4. **Find the instantaneous rate of change at $a=4$:** The instantaneous rate of change is the limit as $h \to 0$ of the difference quotient: $$\lim_{h \to 0} \frac{f(4+h) - f(4)}{h} = \lim_{h \to 0} (2 \cdot 4 + h - 1)$$ Substitute $a=4$ into the simplified difference quotient: $$= \lim_{h \to 0} (8 + h - 1) = \lim_{h \to 0} (7 + h)$$ Taking the limit as $h$ approaches zero: $$= 7$$ So the instantaneous rate of change at $a=4$ is: $$\boxed{7}$$