1. **Problem Statement:** We have a piecewise function defined on the interval $[-2,2]$: $$ f(x) = \begin{cases} \max\left(\sqrt{4 - x^2}, \sqrt{1 + x^2}\right), & -2 \leq x \leq 0 \\ \min\left(\sqrt{4 - x^2}, \sqrt{1 + x^2}\right), & 0 < x \leq 2 \end{cases} $$ We need to analyze the continuity and differentiability of $f(x)$ on $[-2,2]$. 2. **Understanding the functions inside max and min:** - $g(x) = \sqrt{4 - x^2}$ is the upper half of a circle with radius 2 centered at 0. - $h(x) = \sqrt{1 + x^2}$ is an increasing function starting at 1 when $x=0$. 3. **Check where $g(x)$ and $h(x)$ intersect:** Set $\sqrt{4 - x^2} = \sqrt{1 + x^2}$. Square both sides: $$4 - x^2 = 1 + x^2$$ $$4 - 1 = x^2 + x^2$$ $$3 = 2x^2$$ $$x^2 = \frac{3}{2}$$ $$x = \pm \sqrt{\frac{3}{2}} \approx \pm 1.2247$$ 4. **Behavior on $[-2,0]$:** For $x$ in $[-2,0]$, $f(x) = \max(g(x), h(x))$. - At $x=-2$, $g(-2) = 0$, $h(-2) = \sqrt{1+4} = \sqrt{5} \approx 2.236$ so $f(-2) = 2.236$. - At $x=0$, $g(0) = 2$, $h(0) = 1$, so $f(0) = 2$. - Between $-2$ and $-1.2247$, $h(x) > g(x)$, so $f(x) = h(x)$. - Between $-1.2247$ and $0$, $g(x) > h(x)$, so $f(x) = g(x)$. 5. **Behavior on $(0,2]$:** For $x$ in $(0,2]$, $f(x) = \min(g(x), h(x))$. - At $x=0$, $f(0)$ from the right is $\min(2,1) = 1$. - At $x=2$, $g(2) = 0$, $h(2) = \sqrt{5} \approx 2.236$, so $f(2) = 0$. - Between $0$ and $1.2247$, $g(x) > h(x)$, so $f(x) = h(x)$ (since min of $g,h$ is $h$ here). - Between $1.2247$ and $2$, $g(x) < h(x)$, so $f(x) = g(x)$. 6. **Check continuity at $x=0$:** - Left limit: $\lim_{x \to 0^-} f(x) = f(0) = 2$ (from max branch). - Right limit: $\lim_{x \to 0^+} f(x) = \min(2,1) = 1$. - Since left and right limits differ, $f$ is **not continuous at $x=0$**. 7. **Check continuity at $x=\pm 1.2247$:** At these points, $g(x) = h(x)$, so max and min switch between $g$ and $h$ smoothly. Both $g$ and $h$ are continuous and equal at these points, so $f$ is continuous there. 8. **Differentiability:** - At $x=0$, $f$ is not continuous, so not differentiable. - At $x=\pm 1.2247$, $f$ switches between $g$ and $h$. Check derivatives from left and right: For $x=1.2247$: - Derivative of $g(x)$: $$g'(x) = \frac{d}{dx} \sqrt{4 - x^2} = \frac{-x}{\sqrt{4 - x^2}}$$ - Derivative of $h(x)$: $$h'(x) = \frac{d}{dx} \sqrt{1 + x^2} = \frac{x}{\sqrt{1 + x^2}}$$ At $x=1.2247$: $$g'(1.2247) = \frac{-1.2247}{\sqrt{4 - (1.2247)^2}} = \frac{-1.2247}{\sqrt{4 - 1.5}} = \frac{-1.2247}{\sqrt{2.5}} \approx -0.7746$$ $$h'(1.2247) = \frac{1.2247}{\sqrt{1 + 1.5}} = \frac{1.2247}{\sqrt{2.5}} \approx 0.7746$$ Since the left and right derivatives differ, $f$ is **not differentiable at $x=1.2247$**. Similarly at $x=-1.2247$: $$g'(-1.2247) = \frac{-(-1.2247)}{\sqrt{4 - 1.5}} = \frac{1.2247}{\sqrt{2.5}} \approx 0.7746$$ $$h'(-1.2247) = \frac{-1.2247}{\sqrt{1 + 1.5}} = \frac{-1.2247}{\sqrt{2.5}} \approx -0.7746$$ Again, derivatives differ, so $f$ is **not differentiable at $x=-1.2247$**. 9. **Summary:** - $f$ is continuous everywhere except at $x=0$. - $f$ is not differentiable at three points: $x=0$, $x=1.2247$, and $x=-1.2247$. **Final conclusion:** - $f$ is **not continuous at one point** ($x=0$). - $f$ is **not differentiable at more than one point** ($x=0$, $x=\pm 1.2247$). Hence, the correct statement is: **"is not differentiable at more than one point"**. --- "slug": "continuity differentiability", "subject": "calculus", "desmos": {"latex": "f(x) = \max(\sqrt{4 - x^2}, \sqrt{1 + x^2}) \text{ for } x \leq 0, \min(\sqrt{4 - x^2}, \sqrt{1 + x^2}) \text{ for } x > 0", "features": {"intercepts": true, "extrema": true}}, "q_count": 1