1. Problem: Find and simplify the difference quotient $$\frac{f(a+h) - f(a)}{h}$$ for the given functions. (i) \( f(x) = 6x - 9 \) 2. Substitute into the difference quotient formula: $$\frac{f(a+h) - f(a)}{h} = \frac{6(a+h) - 9 - (6a - 9)}{h}$$ 3. Simplify numerator: $$6a + 6h - 9 - 6a + 9 = 6h$$ 4. Thus, $$\frac{6h}{h} = 6$$ (iii) \( f(x) = \sin x \) 5. Substitute into the difference quotient: $$\frac{\sin(a+h) - \sin a}{h}$$ 6. This is the difference quotient definition for the derivative of \( \sin x \) at \( a \). No further algebraic simplification is possible without limits. (iv) \( f(x) = x^3 + 2x^2 - 1 \) 7. Substitute: $$\frac{(a+h)^3 + 2(a+h)^2 - 1 - \big(a^3 + 2a^2 - 1\big)}{h}$$ 8. Expand powers: $$\frac{a^3 + 3a^2h + 3ah^2 + h^3 + 2(a^2 + 2ah + h^2) - 1 - a^3 - 2a^2 + 1}{h}$$ 9. Simplify inside numerator: $$a^3 + 3a^2h + 3ah^2 + h^3 + 2a^2 + 4ah + 2h^2 - 1 - a^3 - 2a^2 + 1 = 3a^2h + 3ah^2 + h^3 + 4ah + 2h^2$$ 10. Group terms: $$ = h(3a^2 + 3ah + h^2 + 4a + 2h)$$ 11. Divide by \( h \): $$3a^2 + 3ah + h^2 + 4a + 2h$$ (v) \( f(x) = \cos x \) 12. Substitute: $$\frac{\cos(a+h) - \cos a}{h}$$ 13. This is the difference quotient for \( \cos x \) at \( a \). No further algebraic simplification without limits. Final answers: (i) $$6$$ (iii) $$\frac{\sin(a+h) - \sin a}{h}$$ (difference quotient form) (iv) $$3a^2 + 3ah + h^2 + 4a + 2h$$ (v) $$\frac{\cos(a+h) - \cos a}{h}$$ (difference quotient form)