1. The problem asks to find the value of $k$ where $k = \frac{f(6) - f(1)}{6 - 1}$ for the function $f(x) = 5x - 2$. 2. First, calculate $f(6)$ and $f(1)$: $$f(6) = 5(6) - 2 = 30 - 2 = 28$$ $$f(1) = 5(1) - 2 = 5 - 2 = 3$$ 3. Substitute these values into the formula for $k$: $$k = \frac{28 - 3}{6 - 1} = \frac{25}{5}$$ 4. Simplify the fraction: $$k = \frac{\cancel{25}}{\cancel{5}} = 5$$ 5. The value of $k$ is 5. This represents the average rate of change of the function $f(x)$ between $x=1$ and $x=6$. 6. Next, find the value of $\lim_{h \to 0} \frac{f(7+h) - f(7)}{h}$. 7. Calculate $f(7+h)$ and $f(7)$: $$f(7+h) = 5(7+h) - 2 = 35 + 5h - 2 = 33 + 5h$$ $$f(7) = 5(7) - 2 = 35 - 2 = 33$$ 8. Substitute into the limit expression: $$\lim_{h \to 0} \frac{(33 + 5h) - 33}{h} = \lim_{h \to 0} \frac{5h}{h}$$ 9. Simplify the fraction: $$\lim_{h \to 0} \frac{\cancel{5h}}{\cancel{h}} = \lim_{h \to 0} 5 = 5$$ 10. The value of the limit is 5, which is the instantaneous rate of change (derivative) of $f$ at $x=7$. 11. Finally, find the value of $\lim_{x \to -1} \frac{f(x) - f(-1)}{x - (-1)}$. 12. Calculate $f(x)$ and $f(-1)$: $$f(x) = 5x - 2$$ $$f(-1) = 5(-1) - 2 = -5 - 2 = -7$$ 13. Substitute into the limit expression: $$\lim_{x \to -1} \frac{(5x - 2) - (-7)}{x + 1} = \lim_{x \to -1} \frac{5x - 2 + 7}{x + 1} = \lim_{x \to -1} \frac{5x + 5}{x + 1}$$ 14. Factor the numerator: $$\lim_{x \to -1} \frac{5(x + 1)}{x + 1}$$ 15. Simplify the fraction: $$\lim_{x \to -1} \frac{5\cancel{(x + 1)}}{\cancel{(x + 1)}} = \lim_{x \to -1} 5 = 5$$ 16. The value of this limit is also 5, which is the instantaneous rate of change of $f$ at $x = -1$. **Final answers:** - a) $k = 5$ - b) $\lim_{h \to 0} \frac{f(7+h) - f(7)}{h} = 5$ - c) $\lim_{x \to -1} \frac{f(x) - f(-1)}{x - (-1)} = 5$