Key Concepts and Formulas

• Function Composition: For two functions $f$ and $g$, the composite function $(g \circ f)(x)$ is defined as $g(f(x))$. To evaluate this, we first find the value of $f(x)$ and then use this value as the input for the function $g$.
• Definition of Absolute Value: $|a| = a$ if $a \geq 0$, and $|a| = -a$ if $a < 0$.
• Continuity of a Function: A function $h(x)$ is continuous at a point $x=c$ if:
  1. $h(c)$ is defined.
  2. $\lim_{x \to c} h(x)$ exists.
  3. $\lim_{x \to c} h(x) = h(c)$. For a piecewise function, we need to check continuity at the points where the definition changes. We do this by comparing the left-hand limit, the right-hand limit, and the function value at that point.
• Differentiability of a Function: A function $h(x)$ is differentiable at a point $x=c$ if the limit of the difference quotient exists: $\lim_{h \to 0} \frac{h(c+h) - h(c)}{h}$. For a piecewise function, differentiability at the points where the definition changes requires the left-hand derivative and the right-hand derivative to be equal.
  • Left-hand derivative: $\lim_{x \to c^-} \frac{h(x) - h(c)}{x-c}$
  • Right-hand derivative: $\lim_{x \to c^+} \frac{h(x) - h(c)}{x-c}$

Step-by-Step Solution

Step 1: Analyze and rewrite the function $f(x)$ in a piecewise form. The function $f(x)$ is defined as: $f(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ |x-1|, & x \geq 0\end{array}\right.$ We need to expand the absolute value term $|x-1|$ for $x \geq 0$.

• If $0 \leq x < 1$, then $x-1 < 0$, so $|x-1| = -(x-1) = 1-x$.
• If $x \geq 1$, then $x-1 \geq 0$, so $|x-1| = x-1$. Therefore, $f(x)$ can be written as: $f(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ 1-x, & 0 \leq x < 1 \\ x-1, & x \geq 1\end{array}\right.$

Step 2: Analyze and rewrite the function $g(x)$. The function $g(x)$ is defined as: $g(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ 1, & x \geq 0\end{array}\right.$ This function is already in a clear piecewise form.

Step 3: Determine the composite function $(g \circ f)(x) = g(f(x))$. We need to consider the definition of $g(x)$ based on the value of its input, which is $f(x)$. The definition of $g$ changes at $f(x) = 0$.

We need to find the values of $x$ for which $f(x) < 0$ and $f(x) \geq 0$.

Case 1: $f(x) < 0$ From the definition of $f(x)$:

• If $x < 0$, then $f(x) = x+1$. So, $x+1 < 0 \implies x < -1$.
• If $0 \leq x < 1$, then $f(x) = 1-x$. So, $1-x < 0 \implies x > 1$. This contradicts $0 \leq x < 1$, so there are no solutions in this interval.
• If $x \geq 1$, then $f(x) = x-1$. So, $x-1 < 0 \implies x < 1$. This contradicts $x \geq 1$, so there are no solutions in this interval. Thus, $f(x) < 0$ only when $x < -1$. For $x < -1$, $f(x) = x+1$. Since $f(x) < 0$, we use the first case of $g(x)$: $g(y) = y+1$. So, $g(f(x)) = g(x+1) = (x+1) + 1 = x+2$ for $x < -1$.

Case 2: $f(x) \geq 0$ We need to find the values of $x$ where $f(x) \geq 0$.

• If $x < 0$, then $f(x) = x+1$. So, $x+1 \geq 0 \implies x \geq -1$. Combining with $x < 0$, this gives $-1 \leq x < 0$.
• If $0 \leq x < 1$, then $f(x) = 1-x$. So, $1-x \geq 0 \implies x \leq 1$. Combining with $0 \leq x < 1$, this gives $0 \leq x < 1$.
• If $x \geq 1$, then $f(x) = x-1$. So, $x-1 \geq 0 \implies x \geq 1$. Combining with $x \geq 1$, this gives $x \geq 1$. Thus, $f(x) \geq 0$ for all $x \geq -1$.

Now we apply the definition of $g(x)$ for $f(x) \geq 0$.

• If $-1 \leq x < 0$: Here $f(x) = x+1$. Since $-1 \leq x < 0$, we have $0 \leq x+1 < 1$. So $f(x) \geq 0$. We use the second case of $g(x)$: $g(y) = 1$. Therefore, $g(f(x)) = g(x+1) = 1$ for $-1 \leq x < 0$.
• If $0 \leq x < 1$: Here $f(x) = 1-x$. Since $0 \leq x < 1$, we have $0 < 1-x \leq 1$. So $f(x) \geq 0$. We use the second case of $g(x)$: $g(y) = 1$. Therefore, $g(f(x)) = g(1-x) = 1$ for $0 \leq x < 1$.
• If $x \geq 1$: Here $f(x) = x-1$. Since $x \geq 1$, we have $x-1 \geq 0$. So $f(x) \geq 0$. We use the second case of $g(x)$: $g(y) = 1$. Therefore, $g(f(x)) = g(x-1) = 1$ for $x \geq 1$.

Combining all these, we get the composite function $(g \circ f)(x)$: $(g \circ f)(x) = \left\{\begin{array}{cc}x+2, & x < -1 \\ 1, & -1 \leq x < 0 \\ 1, & 0 \leq x < 1 \\ 1, & x \geq 1\end{array}\right.$ This can be simplified to: $(g \circ f)(x) = \left\{\begin{array}{cc}x+2, & x < -1 \\ 1, & x \geq -1\end{array}\right.$

Step 4: Check for continuity of $(g \circ f)(x)$. The composite function is defined piecewise with a change in definition at $x=-1$. We need to check continuity at $x=-1$.

• Left-hand limit at $x=-1$: $\lim_{x \to -1^-} (g \circ f)(x) = \lim_{x \to -1^-} (x+2) = -1+2 = 1$.
• Right-hand limit at $x=-1$: $\lim_{x \to -1^+} (g \circ f)(x) = \lim_{x \to -1^+} 1 = 1$.
• Function value at $x=-1$: $(g \circ f)(-1) = 1$ (from the second part of the definition). Since the left-hand limit, right-hand limit, and the function value are all equal to 1, the function $(g \circ f)(x)$ is continuous at $x=-1$. For $x < -1$, $(g \circ f)(x) = x+2$, which is a linear function and thus continuous. For $x > -1$, $(g \circ f)(x) = 1$, which is a constant function and thus continuous. Therefore, $(g \circ f)(x)$ is continuous everywhere.

Step 5: Check for differentiability of $(g \circ f)(x)$. We need to check differentiability at the point where the definition changes, which is $x=-1$.

• Left-hand derivative at $x=-1$: Let $h(x) = (g \circ f)(x)$. For $x < -1$, $h(x) = x+2$. $h'(x) = \frac{d}{dx}(x+2) = 1$. So, the left-hand derivative is $\lim_{x \to -1^-} h'(x) = 1$. Alternatively, using the limit definition: $\lim_{x \to -1^-} \frac{h(x) - h(-1)}{x - (-1)} = \lim_{x \to -1^-} \frac{(x+2) - 1}{x+1} = \lim_{x \to -1^-} \frac{x+1}{x+1} = \lim_{x \to -1^-} 1 = 1$.

• Right-hand derivative at $x=-1$: For $x > -1$, $h(x) = 1$. $h'(x) = \frac{d}{dx}(1) = 0$. So, the right-hand derivative is $\lim_{x \to -1^+} h'(x) = 0$. Alternatively, using the limit definition: $\lim_{x \to -1^+} \frac{h(x) - h(-1)}{x - (-1)} = \lim_{x \to -1^+} \frac{1 - 1}{x+1} = \lim_{x \to -1^+} \frac{0}{x+1} = 0$.

Since the left-hand derivative (1) is not equal to the right-hand derivative (0) at $x=-1$, the function $(g \circ f)(x)$ is not differentiable at $x=-1$.

For $x < -1$, $(g \circ f)(x) = x+2$, which is differentiable everywhere with derivative 1. For $x > -1$, $(g \circ f)(x) = 1$, which is differentiable everywhere with derivative 0.

Therefore, $(g \circ f)(x)$ is continuous everywhere but not differentiable at $x=-1$.

Common Mistakes & Tips

• Incorrectly expanding absolute values: Ensure you correctly define $|x-1|$ for different intervals of $x$. For example, $|x-1| = 1-x$ for $x < 1$ and $|x-1| = x-1$ for $x \geq 1$.
• Forgetting to check continuity/differentiability at the boundary points: The critical points for continuity and differentiability are where the definition of the piecewise function changes. For $(g \circ f)(x)$, this is where $f(x)$ changes its behavior or where the input to $g$ crosses its critical point (which is 0).
• Confusing the input for $g$ with $x$: When calculating $g(f(x))$, remember that the condition for $g$ (i.e., input $< 0$ or input $\geq 0$) applies to $f(x)$, not directly to $x$.

Summary

We first explicitly defined the piecewise function $f(x)$ by resolving the absolute value. Then, we computed the composite function $g(f(x))$ by considering the different cases for the input of $g$ based on the values of $f(x)$. We found that $(g \circ f)(x)$ is a piecewise function that is equal to $x+2$ for $x < -1$ and $1$ for $x \geq -1$. By checking the limits and function values at $x=-1$, we confirmed that the composite function is continuous everywhere. Finally, by comparing the left-hand and right-hand derivatives at $x=-1$, we determined that the function is not differentiable at this single point.

The final answer is $\boxed{A}$.