Consider the function f defined on the domain [\frac{2}{3}, \infty) by the equation f(x) =… — ACT

Consider the function $f$ defined on the domain $[\frac{2}{3}, \infty)$ by the equation $f(x) = 2 + \sqrt{3x - 2}$ . If $f^{-1}$ represents the inverse of $f$ , what is the only real value of $x$ for which $f(x) = f^{-1}(x)$ ?

Answer: 6

Answer

The only real value of

x

for which

f(x) = f^{-1}(x)

is 6.

For a strictly increasing function, the intersection of

f(x)

and

f^{-1}(x)

must occur on the line

y = x

. Equating

f(x) = x

gives

2 + \sqrt{3x - 2} = x

. Isolating the radical term yields

\sqrt{3x - 2} = x - 2

. Squaring both sides produces

3x - 2 = x^2 - 4x + 4

, which simplifies to the quadratic equation

x^2 - 7x + 6 = 0

. Factoring this equation gives

(x - 6)(x - 1) = 0

, yielding the solutions

x = 6

and

x = 1

. Substituting these back into the original equation shows that

x = 6

is a valid solution (

2 + \sqrt{16} = 6

), whereas

x = 1

is extraneous because it results in

2 + 1 = 3 \neq 1

. Furthermore, the domain of

f^{-1}

is the range of

f

, which is

[2, \infty)

, meaning

f^{-1}(1)

is undefined. Thus, the only real solution is 6.

Step-by-Step Solution

Equate the function to

x

using properties of increasing functions and their inverses.

Since

f(x)

is strictly increasing on its domain

[\frac{2}{3}, \infty)

, the graph of

f(x)

and the graph of its inverse

f^{-1}(x)

can only intersect on the line of symmetry

y = x

. Therefore, the equation

f(x) = f^{-1}(x)

is equivalent to

f(x) = x

This simplifies the relation by eliminating the need to solve a high-degree polynomial equation derived from direct composition or substitution.

Set up the equation

f(x) = x

and isolate the radical term.

2 + \sqrt{3x - 2} = x \implies \sqrt{3x - 2} = x - 2

Isolating the square root term is a necessary prerequisite step before squaring both sides.

Square both sides and rewrite the equation as a standard quadratic equation.

3x - 2 = (x - 2)^2 \implies 3x - 2 = x^2 - 4x + 4 \implies x^2 - 7x + 6 = 0

Squaring eliminates the radical to yield a standard quadratic equation that can be solved analytically.

Solve the quadratic equation by factoring.

(x - 6)(x - 1) = 0 \implies x = 6 \text{ or } x = 1

Factoring the quadratic trinomial yields the two candidate values for

x

Check the candidate solutions in the original equation to eliminate extraneous roots.

For

x = 1

2 + \sqrt{3(1)-2} = 3 \neq 1

(extraneous). For

x = 6

2 + \sqrt{3(6)-2} = 2 + 4 = 6

(valid). Also, the domain of

f^{-1}

is the range of

f

, which is

[2, \infty)

, thus excluding

x = 1

Squaring both sides can introduce extraneous roots. We must verify that the solutions satisfy the original radical equation and fall within the domains of both

f

and

f^{-1}

Key Concept

Applying the symmetry of inverse functions about the line

y=x

to solve composition-based equations, while rigorously accounting for domain restrictions and extraneous roots.

Alternative Method

Find the algebraic formula for

f^{-1}(x)

by setting

y = 2 + \sqrt{3x-2}

. Subtracting 2 and squaring both sides gives

(y - 2)^2 = 3x - 2

for

y \geq 2

. Solving for

x

yields

f^{-1}(x) = \frac{(x - 2)^2 + 2}{3}

for

x \geq 2

. Equating

f(x) = f^{-1}(x)

results in

2 + \sqrt{3x - 2} = \frac{(x - 2)^2 + 2}{3}

. Multiplying by 3 and isolating the radical term gives

3\sqrt{3x - 2} = x^2 - 4x

. Squaring both sides results in a fourth-degree polynomial equation:

9(3x - 2) = (x^2 - 4x)^2 \implies x^4 - 8x^3 + 16x^2 - 27x + 18 = 0

. This factors into

(x - 6)(x - 1)(x^2 - x + 3) = 0

. Since the domain of

f^{-1}(x)

is restricted to

x \geq 2

, the root

x = 1

is rejected, and the quadratic factor

x^2 - x + 3 = 0

has no real roots, leaving

x = 6

as the unique real solution.

Estimated Time:3m 0s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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