The functions f and p are defined by \(\mathrm{f(x) = \frac{2}{3}x - 4}\) and \(\mathrm{p(x) = 9x + \frac{5}{2}}\). Which...

1. TRANSLATE the composition notation

• Given information:
  • \(\mathrm{f(x) = \frac{2}{3}x - 4}\)
  • \(\mathrm{p(x) = 9x + \frac{5}{2}}\)
  • Find \(\mathrm{f(p(x))}\)

• What this means: Substitute the entire expression for \(\mathrm{p(x)}\) wherever you see \(\mathrm{x}\) in \(\mathrm{f(x)}\)

2. TRANSLATE the substitution setup

• Replace \(\mathrm{x}\) in \(\mathrm{f(x)}\) with \(\mathrm{p(x)}\):
  \(\mathrm{f(p(x)) = \frac{2}{3}(9x + \frac{5}{2}) - 4}\)

• Now we have our expression to work with

3. SIMPLIFY using the distributive property

• Distribute \(\mathrm{\frac{2}{3}}\) to both terms inside the parentheses:
  \(\mathrm{\frac{2}{3}(9x) = 6x}\)
  \(\mathrm{\frac{2}{3} \cdot \frac{5}{2} = \frac{10}{6} = \frac{5}{3}}\)

• Expression becomes: \(\mathrm{6x + \frac{5}{3} - 4}\)

4. SIMPLIFY by combining constants

• Combine \(\mathrm{\frac{5}{3}}\) and \(\mathrm{-4}\):
  \(\mathrm{-4 = -\frac{12}{3}}\)
  Convert -4 to thirds:
  Add: \(\mathrm{\frac{5}{3} + (-\frac{12}{3}) = -\frac{7}{3}}\)

• Final expression: \(\mathrm{6x - \frac{7}{3}}\)

Answer: C. \(\mathrm{6x - \frac{7}{3}}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students make fraction arithmetic errors when combining \(\mathrm{\frac{5}{3} - 4}\)

Many students correctly set up the substitution and distribute properly, but then struggle with converting -4 to thirds. They might calculate \(\mathrm{\frac{5}{3} - 4}\) incorrectly, perhaps getting \(\mathrm{-\frac{11}{3}}\) or making other arithmetic mistakes.

This may lead them to select Choice A (\(\mathrm{6x - \frac{17}{6}}\)) or get confused and guess.

Second Most Common Error:

Incomplete SIMPLIFY execution: Students don't finish combining the constant terms

Some students get to \(\mathrm{6x + \frac{5}{3} - 4}\) but then don't combine the constants, leaving their answer as \(\mathrm{6x + \frac{5}{3} - 4}\) or incorrectly simplifying it.

This may lead them to select Choice B (\(\mathrm{6x - 4}\)) if they ignore the \(\mathrm{\frac{5}{3}}\) term entirely.

The Bottom Line:

Function composition itself isn't complicated - it's just substitution. The challenge lies in careful algebraic manipulation, especially with fractions. Students need to methodically distribute, then convert all constants to common denominators before combining.