Analyze Function Continuity and Differentiability

The given function is $f(x) = |x| + |x - 1|.$ We need to determine its continuity and differentiability for all real values of x.

Continuity Analysis

The function $f(x)$ is a sum of two absolute value functions, $|x|$ and $|x - 1|$. Both $|x|$ and $|x - 1|$ are continuous functions for all real x. Since the sum of two continuous functions is always continuous, $f(x)$ is continuous for all real values of x.

Differentiability Analysis

Absolute value functions typically have points where they are not differentiable (sharp corners). The points where the argument of the absolute value becomes zero are potential points of non-differentiability. For $f(x)$, these points are $x=0$ (from $|x|$) and $x=1$ (from $|x - 1|$).

Let's define $f(x)$ as a piecewise function:

• For $x < 0$: $f(x) = -x + -(x - 1) = -2x + 1$
• For $0 \leq x < 1$: $f(x) = x + -(x - 1) = 1$
• For $x \geq 1$: $f(x) = x + (x - 1) = 2x - 1$

Now, we check differentiability at the critical points:

Checking Differentiability at x = 0

• Left-hand derivative: $f'(0^-) = \frac{d}{dx}(-2x + 1) \big|_{x=0} = -2$
• Right-hand derivative: $f'(0^+) = \frac{d}{dx}(1) \big|_{x=0} = 0$

Since $f'(0^-) \neq f'(0^+)$, the function is not differentiable at $x = 0$.

Checking Differentiability at x = 1

• Left-hand derivative: $f'(1^-) = \frac{d}{dx}(1) \big|_{x=1} = 0$
• Right-hand derivative: $f'(1^+) = \frac{d}{dx}(2x - 1) \big|_{x=1} = 2$

Since $f'(1^-) \neq f'(1^+)$, the function is not differentiable at $x = 1$.

Conclusion

The function $f(x) = |x| + |x - 1|$ is continuous everywhere but fails to be differentiable at exactly two points, $x = 0$ and $x = 1$. Therefore, the correct statement is that the function is continuous but not differentiable at two points.