Problem: Compute the difference quotient $\frac{f(a+h)-f(a)}{h}$ for the given functions and simplify. 1. For $f(x)=6x-9$. Substitute: $f(a+h)=6(a+h)-9$ and $f(a)=6a-9$. Numerator: $f(a+h)-f(a)=6(a+h)-9-(6a-9)$. Simplify numerator: $6a+6h-9-6a+9=6h$. Divide by $h$: $\frac{f(a+h)-f(a)}{h}=\frac{6h}{h}=6$. Conclusion: The difference quotient is $6$. 2. For $f(x)=\sin x$. Substitute: $f(a+h)=\sin(a+h)$ and $f(a)=\sin a$. Numerator: $f(a+h)-f(a)=\sin(a+h)-\sin a$. Use the identity $\sin u-\sin v=2\cos\left(\frac{u+v}{2}\right)\sin\left(\frac{u-v}{2}\right)$ to write the numerator as $2\cos\left(a+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)$. Divide by $h$: $\frac{f(a+h)-f(a)}{h}=\cos\left(a+\frac{h}{2}\right)\cdot\frac{2\sin\left(\frac{h}{2}\right)}{h}$. This is a standard simplified form useful for limits as $h\to 0$. 3. For $f(x)=x^3+2x^2-1$. Substitute: $f(a+h)=(a+h)^3+2(a+h)^2-1$ and $f(a)=a^3+2a^2-1$. Expand $f(a+h)$: $(a+h)^3=a^3+3a^2h+3ah^2+h^3$ and $2(a+h)^2=2(a^2+2ah+h^2)=2a^2+4ah+2h^2$. So $f(a+h)=a^3+3a^2h+3ah^2+h^3+2a^2+4ah+2h^2-1$. Numerator: $f(a+h)-f(a)=\bigl(a^3+3a^2h+3ah^2+h^3+2a^2+4ah+2h^2-1\bigr)-(a^3+2a^2-1)$. Cancel common terms to get $3a^2h+4ah+3ah^2+2h^2+h^3$. Factor out $h$: $h\left(3a^2+4a+h(3a+2)+h^2\right)$. Divide by $h$: $\frac{f(a+h)-f(a)}{h}=3a^2+4a+h(3a+2)+h^2$. This is the simplified difference quotient. 4. For $f(x)=\cos x$. Substitute: $f(a+h)=\cos(a+h)$ and $f(a)=\cos a$. Numerator: $f(a+h)-f(a)=\cos(a+h)-\cos a$. Use identity $\cos u-\cos v=-2\sin\left(\frac{u+v}{2}\right)\sin\left(\frac{u-v}{2}\right)$ to get $-2\sin\left(a+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)$. Divide by $h$: $\frac{f(a+h)-f(a)}{h}=-\sin\left(a+\frac{h}{2}\right)\cdot\frac{2\sin\left(\frac{h}{2}\right)}{h}$. This is a common form for taking the limit as $h\to 0$.