The function \(\mathrm{F(x) = \frac{9}{5}(x - 273.15) + 32}\) gives the temperature, in degrees Fahrenheit, that corresponds to a temperature...

1. TRANSLATE the problem information

• Given information:
  • Function \(\mathrm{F(x) = \frac{9}{5}(x - 273.15) + 32}\) converts kelvins to Fahrenheit
  • Temperature increased by \(\mathrm{2.10}\) kelvins
  • Need to find the increase in degrees Fahrenheit

2. INFER the mathematical approach

• Key insight: We don't need to know the original temperature!
• We need to find how much \(\mathrm{F(x)}\) increases when \(\mathrm{x}\) increases by \(\mathrm{2.10}\)
• This means calculating \(\mathrm{F(x + 2.10) - F(x)}\)

3. SIMPLIFY the difference calculation

• Original temperature in Fahrenheit: \(\mathrm{F(x) = \frac{9}{5}(x - 273.15) + 32}\)
• New temperature in Fahrenheit: \(\mathrm{F(x + 2.10) = \frac{9}{5}((x + 2.10) - 273.15) + 32}\)
• The increase is:
  \(\mathrm{F(x + 2.10) - F(x) = [\frac{9}{5}(x - 273.15 + 2.10) + 32] - [\frac{9}{5}(x - 273.15) + 32]}\)
• Notice the \(\mathrm{+32}\) terms cancel:
  \(\mathrm{= \frac{9}{5}(x - 273.15 + 2.10) - \frac{9}{5}(x - 273.15)}\)
  \(\mathrm{= \frac{9}{5}[(x - 273.15 + 2.10) - (x - 273.15)]}\)
  \(\mathrm{= \frac{9}{5}(2.10)}\)

4. Calculate the final result

• \(\mathrm{\frac{9}{5} \times 2.10 = 1.8 \times 2.10 = 3.78}\) (use calculator if needed)

Answer: A. 3.78

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students try to find \(\mathrm{F(2.10)}\) directly instead of recognizing they need the difference \(\mathrm{F(x + 2.10) - F(x)}\).

They calculate:
\(\mathrm{F(2.10) = \frac{9}{5}(2.10 - 273.15) + 32}\)
\(\mathrm{= \frac{9}{5}(-271.05) + 32}\)
\(\mathrm{= -487.89 + 32}\)
\(\mathrm{= -455.89}\)

Since this gives a negative result that doesn't match any answer choice, they get confused and may guess, or they might focus on the magnitude and select Choice C (487.89).

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up \(\mathrm{F(x + 2.10) - F(x)}\) but fail to recognize that the constant terms (\(\mathrm{-273.15}\) and \(\mathrm{+32}\)) will cancel out.

They might try to work with specific values of \(\mathrm{x}\) or get overwhelmed by the algebra, leading to calculation errors. Some might incorrectly add the constant \(\mathrm{32}\) to their result of \(\mathrm{3.78}\), leading them to select Choice B (35.78).

The Bottom Line:

This problem tests whether students understand that linear functions have a constant rate of change. The beautiful insight is that the conversion constants (\(\mathrm{-273.15}\) and \(\mathrm{+32}\)) completely cancel out when finding the increase, leaving only the scaling factor \(\mathrm{\frac{9}{5}}\) times the increase amount.