Question:

We analyze the differentiability of $|f(x)|$ for the given function $f(x)$. The function $f(x)$ is defined as: $$f(x) = \begin{cases} x^2 + 1, & x < 0 \\ 2 - x, & x\geq 0 \end{cases}$$ **Step 1: Determine the piecewise definition of $|f(x)|$.** For $x < 0$: $x^2 + 1 > 0$, so $|f(x)| = |x^2 + 1| = x^2 + 1$. For $x \geq 0$: $f(x) = 2 - x$. We consider two sub-cases: * If $0 \leq x < 2$: $2 - x > 0$, so $|f(x)| = |2 - x| = 2 - x$. * If $x \geq 2$: $2 - x \leq 0$, so $|f(x)| = |2 - x| = -(2 - x) = x - 2$. Combining these, the piecewise definition of $|f(x)|$ is: $$|f(x)| = \begin{cases} x^2 + 1, & x < 0 \\ 2 - x, & 0 \leq x < 2 \\ x - 2, & x \geq 2 \end{cases}$$ **Step 2: Analyze differentiability at critical points.** The potential points of non-differentiability are where the definition of $|f(x)|$ changes, which are $x=0$ and $x=2$. We also examine $x=1$ as it is a point of interest. **Differentiability at $x=0$:** For $|f(x)|$ to be differentiable at $x=0$, it must first be continuous at $x=0$. The left-hand limit at $x=0$ is: $$\lim_{x \to 0^-} |f(x)| = \lim_{x \to 0^-} (x^2 + 1) = 0^2 + 1 = 1$$ The right-hand limit at $x=0$ is: $$\lim_{x \to 0^+} |f(x)| = \lim_{x \to 0^+} (2 - x) = 2 - 0 = 2$$ Since $\lim_{x \to 0^-} |f(x)| \neq \lim_{x \to 0^+} |f(x)|$, $|f(x)|$ is not continuous at $x=0$. Therefore, $|f(x)|$ is not differentiable at $x=0$. **Differentiability at $x=1$:** At $x=1$, the function $|f(x)|$ is defined by $2-x$, as $0 \leq 1 < 2$. The derivative of $2-x$ is $-1$. Since $x=1$ lies within an open interval where $|f(x)| = 2-x$, and $2-x$ is a polynomial, $|f(x)|$ is differentiable at $x=1$. The derivative at $x=1$ is $|f|'(1) = -1$. **Differentiability at $x=2$:** First, check for continuity at $x=2$. The value of the function at $x=2$ is $|f(2)| = 2 - 2 = 0$. The left-hand limit at $x=2$ is: $$\lim_{x \to 2^-} |f(x)| = \lim_{x \to 2^-} (2 - x) = 2 - 2 = 0$$ The right-hand limit at $x=2$ is: $$\lim_{x \to 2^+} |f(x)| = \lim_{x \to 2^+} (x - 2) = 2 - 2 = 0$$ Since the left-hand limit, right-hand limit, and function value are all equal, $|f(x)|$ is continuous at $x=2$. Next, check the left-hand and right-hand derivatives at $x=2$. The left-hand derivative at $x=2$ is: $$|f|'(2^-) = \lim_{x \to 2^-} \frac{d}{dx}(2-x) = -1$$ The right-hand derivative at $x=2$ is: $$|f|'(2^+) = \lim_{x \to 2^+} \frac{d}{dx}(x-2) = 1$$ Since $|f|'(2^-) \neq |f|'(2^+)$, $|f(x)|$ is not differentiable at $x=2$. **Conclusion:** Based on the analysis, $|f(x)|$ is differentiable at $x=1$. It is not differentiable at $x=0$ or $x=2$.