Question

Consider the piecewise function \[ f(x) = \begin{cases} 3 - x, & x \leq 0 \\ x^2 + 2, & x > 0 \end{cases} \] Evaluate \(f(-3)\).

• A. 6
• B. 0
• C. 11
• D. -7

Hint:

Since \(-3 \leq 0\), use the branch \(f(x) = 3 - x\), not \(x^2+2\), which only applies for \(x > 0\).

Answer 1

The Correct Option is A

Step 1: Decide which branch applies.
The function has two branches: \(3 - x\) for \(x \leq 0\), and \(x^2 + 2\) for \(x > 0\).
We need \(f(-3)\), and since \(-3 \leq 0\), the first branch, \(3 - x\), is the one that applies.

Step 2: Substitute \(x = -3\) into the correct branch.
\[ f(-3) = 3 - (-3) \]
Subtracting a negative number is the same as adding its positive value, so this becomes:
\[ f(-3) = 3 + 3 = 6 \]

Step 3: Check why the other options are traps.
Option C (11) is what you get if you wrongly use the second branch: \((-3)^2 + 2 = 9 + 2 = 11\). This branch is only valid for \(x > 0\), and \(-3\) does not satisfy that condition, so 11 is incorrect here.
Option B (0) comes from a sign slip, treating \(3 - (-3)\) as if it were \(3 - 3 = 0\), forgetting that subtracting a negative flips the sign.
Option D (\(-7\)) does not match a correct substitution into either branch and is simply a distractor.

Final Answer:
Since \(-3 \leq 0\), we use \(3 - x\), giving \(f(-3) = 6\).
\[ \boxed{f(-3) = 6} \]