1. **State the problem:** Find the slope of the tangent line at $x=2$ for the functions $f(x) = 4x - x^2$ and $g(x) = \frac{1}{3x - 7}$ using the definition of the derivative. 2. **Recall the definition of the derivative:** $$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ This gives the slope of the tangent line to the curve at $x=a$. --- ### For $f(x) = 4x - x^2$ at $x=2$: 3. Compute $f(2+h)$: $$f(2+h) = 4(2+h) - (2+h)^2 = 8 + 4h - (4 + 4h + h^2) = 8 + 4h - 4 - 4h - h^2 = 4 - h^2$$ 4. Compute the difference quotient: $$\frac{f(2+h) - f(2)}{h} = \frac{(4 - h^2) - (4 - 4)}{h} = \frac{4 - h^2 - 4 + 4}{h} = \frac{4 - h^2}{h}$$ Note $f(2) = 4(2) - 2^2 = 8 - 4 = 4$. 5. Simplify the difference quotient: $$\frac{4 - h^2}{h} = \frac{4}{h} - h$$ 6. Take the limit as $h \to 0$: $$\lim_{h \to 0} \left( \frac{4}{h} - h \right)$$ This expression is incorrect because the previous step miscalculated the difference quotient. Let's correct step 4. **Correction:** $$\frac{f(2+h) - f(2)}{h} = \frac{(4 - h^2) - 4}{h} = \frac{4 - h^2 - 4}{h} = \frac{-h^2}{h} = -h$$ 7. Now take the limit: $$f'(2) = \lim_{h \to 0} -h = 0$$ So the slope of the tangent line to $f$ at $x=2$ is $0$. --- ### For $g(x) = \frac{1}{3x - 7}$ at $x=2$: 8. Compute $g(2+h)$: $$g(2+h) = \frac{1}{3(2+h) - 7} = \frac{1}{6 + 3h - 7} = \frac{1}{-1 + 3h}$$ 9. Compute the difference quotient: $$\frac{g(2+h) - g(2)}{h} = \frac{\frac{1}{-1 + 3h} - \frac{1}{-1}}{h} = \frac{\frac{1}{-1 + 3h} + 1}{h}$$ 10. Find a common denominator inside the numerator: $$\frac{1}{-1 + 3h} + 1 = \frac{1}{-1 + 3h} + \frac{-1 + 3h}{-1 + 3h} = \frac{1 -1 + 3h}{-1 + 3h} = \frac{3h}{-1 + 3h}$$ 11. Substitute back: $$\frac{g(2+h) - g(2)}{h} = \frac{\frac{3h}{-1 + 3h}}{h} = \frac{3h}{h(-1 + 3h)} = \frac{3}{-1 + 3h}$$ 12. Take the limit as $h \to 0$: $$g'(2) = \lim_{h \to 0} \frac{3}{-1 + 3h} = \frac{3}{-1} = -3$$ --- **Final answers:** - Slope of tangent to $f(x)$ at $x=2$ is $\boxed{0}$. - Slope of tangent to $g(x)$ at $x=2$ is $\boxed{-3}$.