The function $f: \mathbb{R} \to \mathbb{R}$ defined by

$f(x) = \begin{cases} x^2, & x \ge 1 \\ x, & x < 1 \end{cases}$

is

The function is defined as:

$f(x) = \begin{cases} x^2, & x \ge 1 \\ x, & x < 1 \end{cases}$

The function switches formulas at $x = 1$, so continuity and differentiability need to be checked at this point.

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For continuity at $x = 1$, the left-hand limit, right-hand limit, and function value must all be equal.

$f(1) = 1^2 = 1$

$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$

$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1^2 = 1$

Since all three values equal $1$, the function is continuous at $x = 1$.

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For differentiability at $x = 1$, the left-hand derivative and right-hand derivative must be equal.

For $x < 1$, $f(x) = x$

$f'(x) = 1$

$f'(1^-) = 1$

For $x \ge 1$, $f(x) = x^2$

$f'(x) = 2x$

$f'(1^+) = 2(1) = 2$

Since $1 \neq 2$, the left-hand derivative does not equal the right-hand derivative. The function is not differentiable at $x = 1$.

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The function is continuous but not differentiable at $x = 1$.