For $f(-3)$:

Since $-3 < 0$, use the first formula: $f(-3) = 2(-3) = -6$

For $f(2)$:

Since $2 \geq 0$, use the second formula: $f(2) = 2 + 5 = 7$

Answer: $f(-3) = -6$, $f(2) = 7$

The absolute value changes behavior at the point where the inside equals zero.

$x - 3 = 0$ when $x = 3$

When $x \geq 3$: the inside is positive or zero $|x - 3| = x - 3$

When $x < 3$: the inside is negative $|x - 3| = -(x - 3) = -x + 3$

Answer: $f(x) = \begin{cases} -x + 3 & \text{if } x < 3 \\ x - 3 & \text{if } x \geq 3 \end{cases}$

Check if the function value and limits match at $x = 1$.

From left: (use $x^2 + 1$) $\lim_{x \to 1^-} f(x) = 1^2 + 1 = 2$

From right: (use $3x - 1$) $\lim_{x \to 1^+} f(x) = 3(1) - 1 = 2$

Function value: (use $x^2 + 1$ since $1 \leq 1$) $f(1) = 1^2 + 1 = 2$

Since all three equal 2, the function is continuous at $x = 1$.

Answer: Yes, continuous at $x = 1$