Question 874·Hard·Math

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The function $f$ is defined for all real numbers $x \neq 4$ by

f(x)=\frac{2x+3}{x-4}.

Let $S$ be the set of all real numbers $a$ such that $f\bigl(f(a)\bigr)=a$ . What is the sum of the elements of $S$ ?

For composite rational functions like this, first simplify the composition algebraically: compute $f(f(x))$ in general, watching for domain restrictions so you do not multiply by zero when clearing denominators. Then set this expression equal to $x$ , cross-multiply to remove fractions, and rearrange into a standard quadratic. If the question only asks for the sum of the solutions, use the sum-of-roots shortcut ( $-b$ for $x^2+bx+c=0$ ) instead of solving the quadratic fully; this saves time and reduces algebraic mistakes.

Hints

Hint 1
Think about the composition explicitly
Write out $f(f(a))$ by first letting $y = f(a)$ and then computing $f(y) = \dfrac{2y+3}{y-4}$ . Then substitute $y = \dfrac{2a+3}{a-4}$ .
Hint 2
Simplify the composed function
When you plug $y = \dfrac{2a+3}{a-4}$ into $\dfrac{2y+3}{y-4}$ , carefully combine fractions so that the numerator and denominator each become a single fraction in terms of $a$ , then simplify the overall expression.
Hint 3
Turn the equation into a quadratic
Once you have a simplified expression for $f(f(a))$ in terms of $a$ , set it equal to $a$ and clear the denominator by multiplying both sides by that denominator. Rearrange the resulting equation into the standard quadratic form $a^2 + ba + c = 0$ .
Hint 4
Use the sum-of-roots shortcut
After you have the quadratic equation, you do not need the individual solutions to find the sum of the elements of $S$ . Use the fact that for $x^2 + bx + c = 0$ , the sum of the solutions is $-b$ .

Desmos guide

Step 1
Enter the original function
In the first expression line, type f(x) = (2x+3)/(x-4) so Desmos defines the function $f$ .
Step 2
Define the composed function
In the next line, type g(x) = (2*f(x)+3)/(f(x)-4) so that $g(x)$ represents $f(f(x))$ .
Step 3
Graph and find intersection points
Add another line with y = x. Look at the graph of $y = g(x)$ and $y = x$ and find their intersection points. The $x$ -coordinates of these intersection points are the values of $a$ that satisfy $f(f(a)) = a$ (ignore the vertical asymptotes near $x = 4$ and $x = 9.5$ ).
Step 4
Compute the sum of the solutions
Record the two $x$ -values from the intersection points and add them together (you can type their approximate values in a new expression like a1 + a2). The result of this addition should match one of the answer choices.

Step-by-step

1
Step 1
Set up the equation for the composition
We are given
$f(x)=\frac{2x+3}{x-4}$
for all real $x \neq 4$ . We want all real numbers $a$ such that
$f(f(a))=a.$
So our goal is to:

Find a formula for $f(f(x))$ .

Set $f(f(a)) = a$ and solve for $a$ , remembering that $a \neq 4$ and also $f(a) \neq 4$ (so the inner $f$ is defined).
2
Step 2
Compute the composition $f(f(x))$
Let $y=f(x)$ . Then
$y=\frac{2x+3}{x-4}.$
Now
$f(f(x))=f(y)=\frac{2y+3}{y-4}.$
Replace $y$ with $\dfrac{2x+3}{x-4}$ :

Numerator:

$2y+3 = 2\cdot \frac{2x+3}{x-4} + 3 = \frac{4x+6}{x-4} + 3 = \frac{4x+6 + 3(x-4)}{x-4} = \frac{7x-6}{x-4}.$

Denominator:

$y-4 = \frac{2x+3}{x-4} - 4 = \frac{2x+3 - 4(x-4)}{x-4} = \frac{-2x+19}{x-4} = \frac{19-2x}{x-4}.$
So
$f(f(x)) = \frac{\dfrac{7x-6}{x-4}}{\dfrac{19-2x}{x-4}} = \frac{7x-6}{19-2x},$
with the understanding that $x \neq 4$ and $x \neq \dfrac{19}{2}$ (since $19-2x \neq 0$ ).
Step 3
Set $f(f(a))=a$ and form a quadratic
Now apply the composition to $a$ :
$f(f(a)) = \frac{7a-6}{19-2a}.$
We are given $f(f(a)) = a$ , so
$\frac{7a-6}{19-2a} = a,$
with $a \neq 4$ and $a \neq \dfrac{19}{2}$ .

Clear the fraction by multiplying both sides by $19-2a$ (which we know is not zero for valid $a$ ):
$7a - 6 = a(19-2a).$
Expand the right side and bring all terms to one side:
$7a - 6 = 19a - 2a^2 \\[1.4em] 0 = 19a - 2a^2 - 7a + 6 \\[1.4em] 0 = -2a^2 + 12a + 6.$
Multiply by $-1$ and divide by $2$ to simplify:
$a^2 - 6a - 3 = 0.$
So any valid $a$ satisfying $f(f(a))=a$ must be a solution to this quadratic equation (and must not be $4$ or $\dfrac{19}{2}$ ).
Step 4
Find the sum of the solutions to the quadratic
The equation
$a^2 - 6a - 3 = 0$
is a monic quadratic of the form $a^2 + ba + c = 0$ with $b = -6$ and $c = -3$ .

For any quadratic $x^2 + bx + c = 0$ , the sum of its solutions is $-b$ .

Here, $-b = -(-6)$ , so the sum of the solutions to $a^2 - 6a - 3 = 0$ is
$-(-6) = 6.$
Both solutions are real and are not equal to $4$ or $\dfrac{19}{2}$ , so they are both valid elements of $S$ . Therefore, the sum of the elements of $S$ is $\boxed{6}$ , which corresponds to answer choice C.

Question 874·Hard·Math

Strategy

Hints

Desmos guide

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Desmos guide

Step-by-step

Lessons that teach this