If $f(x) = \frac{5x + 2}{3x - 5}$ and $g(x) = x^2 - 2x - 1$, then the value of $g(f(f(3)))$ is

When we see expressions like g(f(f(3))), we're dealing with function composition. This means we work from the inside out, like peeling an onion:

First: Find f(3)

Then: Find f(f(3))

Finally: Find g(f(f(3)))

Given functions:

$f(x) = \dfrac{5x + 2}{3x - 5}$

$g(x) = x^2 - 2x - 1$

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We substitute x = 3 into f(x):

$f(3) = \dfrac{5(3) + 2}{3(3) - 5}$

$= \dfrac{15 + 2}{9 - 5}$

$= \dfrac{17}{4}$

Always substitute carefully and follow order of operations (PEMDAS/BODMAS).

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Now we substitute x = 17/4 into f(x):

$f\left(\dfrac{17}{4}\right) = \dfrac{5 \cdot \frac{17}{4} + 2}{3 \cdot \frac{17}{4} - 5}$

Numerator calculation:

$5 \cdot \tfrac{17}{4} + 2 = \tfrac{85}{4} + 2$

To add these fractions, convert 2 to fourths: $2 = \tfrac{8}{4}$

$\dfrac{85}{4} + \dfrac{8}{4} = \dfrac{93}{4}$

Denominator calculation:

$3 \cdot \tfrac{17}{4} - 5 = \tfrac{51}{4} - 5$

Convert 5 to fourths: $5 = \tfrac{20}{4}$

$\dfrac{51}{4} - \dfrac{20}{4} = \dfrac{31}{4}$

Final calculation:

$f\left(\dfrac{17}{4}\right) = \dfrac{\frac{93}{4}}{\frac{31}{4}}$

$= \dfrac{93}{4} \times \dfrac{4}{31}$

$= \dfrac{93}{31}$

$= 3$

When dividing fractions, multiply by the reciprocal. The 4's cancel out beautifully!

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Since f(f(3)) = 3, we now find g(3):

$g(3) = 3^2 - 2(3) - 1$

$= 9 - 6 - 1$

$= 2$

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$g(f(f(3))) = 2$

Function composition problems always work from the inside out. Take your time with fraction arithmetic - it's where most mistakes happen!