1. **Stating the problem:** Given $r>0$ and $k<0$, consider the circle $C_r$ centered at the origin with radius $r$, and the function $f_k(x) = k|x|$. We need to verify that $f_k$ is continuous but not differentiable at $x=0$ for any $k$. Then find the two values of $r$ such that $C_r$ and the graph of $f_k$ bound a circular sector in the half-plane $y \leq 0$ with area $\pi$ and perimeter $4 + \pi$. It is given that $r=2$ is the larger of these values. Finally, plot $C_2$ and $f_{-1}$. 2. **Continuity and differentiability of $f_k(x) = k|x|$ at $x=0$:** - The absolute value function $|x|$ is continuous everywhere, so $f_k$ is continuous everywhere. - To check differentiability at $x=0$, compute the left and right derivatives: $$f_k'(0^-) = \lim_{h \to 0^-} \frac{f_k(h) - f_k(0)}{h} = \lim_{h \to 0^-} \frac{k|h| - 0}{h} = \lim_{h \to 0^-} \frac{k(-h)}{h} = \lim_{h \to 0^-} -k = -k$$ $$f_k'(0^+) = \lim_{h \to 0^+} \frac{f_k(h) - f_k(0)}{h} = \lim_{h \to 0^+} \frac{k|h| - 0}{h} = \lim_{h \to 0^+} \frac{kh}{h} = k$$ Since $k<0$, $f_k'(0^-) = -k > 0$ and $f_k'(0^+) = k < 0$, so the left and right derivatives are not equal. Therefore, $f_k$ is not differentiable at $x=0$. 3. **Finding $r$ values for the circular sector with area $\pi$ and perimeter $4 + \pi$:** - The circle $C_r$ has equation $x^2 + y^2 = r^2$. - The function $f_k(x) = k|x|$ is a V-shaped graph with vertex at the origin. - The sector is in the half-plane $y \leq 0$, so the sector is bounded by the circle and the graph of $f_k$ below the $x$-axis. - The perimeter of the sector is the sum of the arc length of the circle segment and the length of the two straight lines from the origin to the intersection points. - The perimeter is given as $4 + \pi$. - The area of the sector is $\pi$. - The length of the two straight lines is $2r$ (since the points lie on the circle radius $r$). - The arc length of the sector is $r\theta$, where $\theta$ is the central angle in radians. - The perimeter condition: $2r + r\theta = 4 + \pi$. - The area of the sector: $\frac{1}{2} r^2 \theta = \pi$. - From area: $\theta = \frac{2\pi}{r^2}$. - Substitute into perimeter: $$2r + r \cdot \frac{2\pi}{r^2} = 4 + \pi \implies 2r + \frac{2\pi}{r} = 4 + \pi$$ Multiply both sides by $r$: $$2r^2 + 2\pi = (4 + \pi) r$$ Rearranged: $$2r^2 - (4 + \pi) r + 2\pi = 0$$ Use quadratic formula: $$r = \frac{4 + \pi \pm \sqrt{(4 + \pi)^2 - 16\pi}}{4}$$ Calculate discriminant: $$(4 + \pi)^2 - 16\pi = 16 + 8\pi + \pi^2 - 16\pi = 16 - 8\pi + \pi^2 = (4 - \pi)^2$$ So: $$r = \frac{4 + \pi \pm (4 - \pi)}{4}$$ Two solutions: - $$r_1 = \frac{4 + \pi + 4 - \pi}{4} = \frac{8}{4} = 2$$ - $$r_2 = \frac{4 + \pi - 4 + \pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2}$$ Thus, the two values of $r$ are $2$ and $\frac{\pi}{2}$. 4. **Plotting $C_2$ and $f_{-1}$:** - Circle $C_2$: $x^2 + y^2 = 4$. - Function $f_{-1}(x) = -|x|$. - The graph of $f_{-1}$ is a V shape opening downward with vertex at origin. - The circle $C_2$ is centered at origin with radius 2. **Final answer:** - $f_k$ is continuous but not differentiable at $x=0$ for any $k$. - The two values of $r$ are $r=2$ and $r=\frac{\pi}{2}$. - $r=2$ is the larger value. - The circle $C_2$ and the graph of $f_{-1}$ are as described.