The function h is defined by \(\mathrm{h(x) = \begin{cases} x + 5 & \text{if } x \geq 0 \\ -x...

1. TRANSLATE the problem information

• Given: \(\mathrm{h(x) = \begin{cases} x + 5 & \text{if } x \geq 0 \\ -x + 3 & \text{if } x \lt 0 \end{cases}}\)
• Find: \(\mathrm{h(-2)}\)
• This means we need to substitute \(\mathrm{x = -2}\) into the appropriate piece of the function

2. INFER which piece of the function to use

• We have \(\mathrm{x = -2}\)
• Check the conditions: Is \(\mathrm{-2 \geq 0}\)? No, \(\mathrm{-2 \lt 0}\)
• Since \(\mathrm{-2 \lt 0}\), we use the second piece: \(\mathrm{h(x) = -x + 3}\)

3. SIMPLIFY by substituting and calculating

• Substitute \(\mathrm{x = -2}\) into \(\mathrm{h(x) = -x + 3}\):
• \(\mathrm{h(-2) = -(-2) + 3}\)
• \(\mathrm{h(-2) = 2 + 3 = 5}\)

Answer: C (5)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students use the wrong piece of the piecewise function by not correctly checking which condition applies to \(\mathrm{x = -2}\).

They might think "since we're dealing with negative numbers, let's use the first piece" or simply choose randomly between the pieces. Using \(\mathrm{h(x) = x + 5}\) gives \(\mathrm{h(-2) = -2 + 5 = 3}\).

This may lead them to select Choice B (3).

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify the right piece but make an arithmetic error with the double negative.

They know to use \(\mathrm{-x + 3}\) but incorrectly think that \(\mathrm{-(-2) = -2}\) instead of \(\mathrm{+2}\). This gives \(\mathrm{h(-2) = -2 + 3 = 1}\).

This may lead them to select Choice A (1).

The Bottom Line:

Piecewise functions require careful attention to domain conditions AND precise arithmetic with negative numbers. Students must systematically check which piece applies before substituting, then execute the algebra carefully to avoid sign errors.