Analyzing the Continuity and Differentiability of f(x) = |x - 1| + x²

The given function is \(f(x) = \vert x - 1 \vert + x^2\), where \(x \in \mathbb{R}\).

The absolute value function \(|x - 1|\) changes its definition depending on the value of \(x - 1\). Specifically:

• If \(x - 1 \ge 0\), which means \(x \ge 1\), then \(\vert x - 1 \vert = x - 1\).
• If \(x - 1 < 0\), which means \(x < 1\), then \(\vert x - 1 \vert = -(x - 1) = 1 - x\).

Therefore, we can rewrite the function \(f(x)\) as a piecewise function:

\[f(x) = \begin{cases} (1 - x) + x^2 = x^2 - x + 1 & \text{if } x < 1 \\ (x - 1) + x^2 = x^2 + x - 1 & \text{if } x \ge 1 \end{cases}\]

Continuity Check at x = 1

For a function to be continuous at a point \(x=c\), the left-hand limit, the right-hand limit, and the function value at \(c\) must all be equal.

Let's check continuity at \(x = 1\):

• Left-hand limit at \(x=1\):
  \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 - x + 1)\)
  Since \(x < 1\), we use the first part of the function definition.
  \(\lim_{x \to 1^-} (x^2 - x + 1) = 1^2 - 1 + 1 = 1\)
• Right-hand limit at \(x=1\):
  \(\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2 + x - 1)\)
  Since \(x \ge 1\), we use the second part of the function definition.
  \(\lim_{x \to 1^+} (x^2 + x - 1) = 1^2 + 1 - 1 = 1\)
• Function value at \(x=1\):
  \(f(1) = 1^2 + 1 - 1 = 1\)
  Using the second part of the definition where \(x \ge 1\).

Since \(\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) = 1\), the function \(f(x)\) is continuous at \(x = 1\).

Differentiability Check at x = 1

For a function to be differentiable at a point \(x=c\), the left-hand derivative and the right-hand derivative at \(c\) must be equal.

First, let's find the derivative of each part of the piecewise function:

• For \(x < 1\), \(f'(x) = \frac{d}{dx}(x^2 - x + 1) = 2x - 1\).
• For \(x > 1\), \(f'(x) = \frac{d}{dx}(x^2 + x - 1) = 2x + 1\).

Now, let's evaluate the left-hand derivative and right-hand derivative at \(x=1\):

• Left-hand derivative at \(x=1\):
  \(LHD = \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} (2x - 1)\)
  Using the derivative for \(x < 1\).
  \(LHD = 2(1) - 1 = 1\)
• Right-hand derivative at \(x=1\):
  \(RHD = \lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} (2x + 1)\)
  Using the derivative for \(x > 1\).
  \(RHD = 2(1) + 1 = 3\)

Since \(LHD \ne RHD\) (1 vs 3), the function \(f(x)\) is not differentiable at \(x = 1\).

Analysis at x = 0

The point \(x = 0\) falls in the range \(x < 1\). For this range, the function is defined as \(f(x) = x^2 - x + 1\). This is a polynomial function.

Polynomial functions are continuous and differentiable for all real numbers. Therefore, \(f(x)\) is continuous and differentiable at \(x = 0\).

Conclusion

Based on our analysis:

• At \(x = 1\), the function \(f(x)\) is continuous but not differentiable.
• At \(x = 0\), the function \(f(x)\) is continuous and differentiable.

Let's review the given options:

1. f(x) is continuous but not differentiable at x = 0. (Incorrect - it's differentiable at x=0)
2. f(x) is continuous but not differentiable at x = 1. (Correct - aligns with our findings)
3. f(x) is differentiable at x = 1. (Incorrect - it is not differentiable at x=1)
4. f(x) is not differentiable at x = 0 and x = 1. (Incorrect - it is differentiable at x=0)

Thus, the correct statement is that \(f(x)\) is continuous but not differentiable at \(x = 1\).