Let f(x) = αx / (x + 1), x ≠ -1. Then, for what value of α is f(f(x)) = x? (a) √2 (b) -√2 (c) 1 (d)

Q: Wait, why do we set the coefficients to zero? Can't we just divide by $x$?

Great question! If we divide by $x$, we lose information. The condition $f(f(x)) = x$ must hold for all $x$ in the domain. If you have a polynomial equation $Ax^2 + Bx = 0$ that is true for all $x$, the only way that happens is if $A=0$ and $B=0$ simultaneously. Dividing by $x$ would only give you a specific point, not the general identity.

Q: Why did we get $\alpha = \pm 1$ from the $x$ term but only $\alpha = -1$ from the $x^2$ term?

Because the condition must satisfy the entire equation. If $\alpha = 1$, the $x^2$ coefficient becomes $1+1=2$, which is not zero. Thus, $\alpha = 1$ fails to make the $x^2$ term vanish. Only $\alpha = -1$ satisfies both requirements, making it the unique solution.

Q: Is there a faster way to do this without all that algebra?

Yes! You could find the inverse function $f^{-1}(x)$ by swapping $x$ and $y$ and solving for $y$. Since $f(f(x)) = x$ implies $f(x) = f^{-1}(x)$, you can equate the original function to its inverse. It's a beautiful shortcut that confirms our algebraic result.

Q: What happens if $x = -1$? The problem says $x \neq -1$. Does that matter?

It matters for the domain of the function, but not for finding the constant $\alpha$. We are looking for the value of $\alpha$ that makes the identity hold wherever the function is defined. We ignore the singularity at $x = -1$ because the identity must hold for all other values in the domain.

Q: Why is $f(f(x)) = x$ called an involution?

An involution is a mathematical function that is its own inverse. Think of it like a light switch: flipping it once changes the state, but flipping it twice returns it to the original state. It's a fundamental concept in symmetry and group theory!

Question

Let $f(x) = \frac{\alpha x}{x + 1}, x 
eq -1$. Then, for what value of $\alpha$ is $f(f(x)) = x$?

Options:

(A) $\sqrt{2}$

(B) $-\sqrt{2}$

(C) $1$

(D) $-1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex machine. You feed it a number $x$, and it transforms that number into a new value, $f(x) = \frac{\alpha x}{x + 1}$.

Now, imagine you take the output of that machine and feed it back into the very same machine. The question asks: for what value of $\alpha$ does this two-step process return the original input $x$ exactly as it was?

This is the essence of an **involution**—a function that acts as its own inverse. Let us embark on this algebraic journey together.

### The Composition

To understand $f(f(x))$, we must treat the function as a template. The rule is simple: take whatever is inside the parentheses, multiply it by $\alpha$, and divide by that same input plus one.

When we nest the function, we replace the $x$ in the denominator and the numerator with the entire expression $\frac{\alpha x}{x + 1}$. It looks daunting, but stay with me:

$$f(f(x)) = \frac{\alpha \left( \frac{\alpha x}{x + 1} \right)}{\left( \frac{\alpha x}{x + 1} \right) + 1}$$

### Taming the Fraction

I know this looks like a mess of fractions, but let's breathe. We simplify the numerator first: $\frac{\alpha^2 x}{x + 1}$.

Now, look at the denominator. To add $1$ to $\frac{\alpha x}{x + 1}$, we need a common denominator, which is $(x + 1)$. This gives us $\frac{\alpha x + x + 1}{x + 1}$, or more elegantly, $\frac{(\alpha + 1)x + 1}{x + 1}$.

When we stack these, the $(x + 1)$ terms in the denominators of both the top and bottom fractions cancel out beautifully. This is the moment of clarity where the complexity collapses into:

$$f(f(x)) = \frac{\alpha^2 x}{(\alpha + 1)x + 1}$$

### The Identity Constraint

We are told that $f(f(x)) = x$. So, we set our simplified expression equal to $x$:

$$\frac{\alpha^2 x}{(\alpha + 1)x + 1} = x$$

Cross-multiplying gives us $\alpha^2 x = x [(\alpha + 1)x + 1]$. Expanding the right side, we get $\alpha^2 x = (\alpha + 1)x^2 + x$.

To solve this, we bring everything to one side to form a polynomial equation:

$$(\alpha + 1)x^2 + (1 - \alpha^2)x = 0$$

### The Final Revelation

For this equation to hold true for all $x$, the coefficients of the polynomial must be zero. If the coefficient of $x^2$ is not zero, the equation would only be true for specific values of $x$, not as an identity.

Thus, we require $\alpha + 1 = 0$, which gives us $\alpha = -1$. Checking the $x$ coefficient, $1 - \alpha^2 = 0$ also yields $\alpha = \pm 1$.

Since $\alpha = -1$ is the only value that satisfies both conditions, we have found our answer. It is a beautiful result—a simple constant that transforms a complex rational function into a perfect mirror of itself.

**The final value is $\alpha = -1$.**

Animated Solution for Mathematics - Functions: Let $f (x) = \frac{α x}{x + 1}, x \neq = - 1$ . Then, for what value of $α$ is $f (f (x)) = x$ ?

Select Answer:

Visualized Solution

Understanding the Goal

Defining the Composition $f (f (x))$

Substituting the Expression

Simplifying the Numerator

Simplifying the Denominator

Combining and Canceling

Setting the Identity

Cross-Multiplication

Analyzing the Polynomial

Solving for $α$

Final Conclusion

The Sigma Insight: Composite Functions

Analyzing the Setup

The Composition

Taming the Fraction

The Identity Constraint

The Final Revelation

Similar Questions

Let $f (x) = \frac{x - 1}{x + 1}, x \in R - {0, - 1, 1}$ . If $f^{n + 1} (x) = f (f^{n} (x))$ for all $n \in N$ , then $f^{6} (6) + f^{7} (7)$ is equal to:

Let $f : R \to R$ be a function defined by $f (x) = [2 (1 - \frac{x ^{25}}{2}) (2 + x^{25})]^{\frac{1}{50}}$ . If the function $g (x) = f (f (f (x))) + f (f (x))$ , then the greatest integer less than or equal to $g (1)$ is ______.

For a suitably chosen real constant $a$ , let a function, $f : R - {- a} \to R$ be defined by $f (x) = \frac{a - x}{a + x}$ . Further suppose that for any real number $x \neq = - a$ and $f (x) \neq = - a$ , $(f o f) (x) = x$ . Then $f (- \frac{1}{2})$ is equal to

Consider the function $f : R \to R$ defined by $f (x) = \frac{2 x}{1 + 9 x ^{2}}$ . If the composition of $f, 10 times (f \circ f \circ f \circ \dots \circ f) (x) = \frac{2 ^{10} x}{1 + 9 α x ^{2}}$ , then the value of $3 α + 1$ is equal to ______

For $x \in R$ , two real valued functions $f (x)$ and $g (x)$ are such that, $g (x) = x + 1$ and $f (g (x)) = x + 3 - x$ . Then $f (0)$ is equal to

If $f (x) = \frac{( t a n 1 ^{\circ} ) ^{x} + l o g _{e} ( 123 )}{x l o g _{e} ( 1234 ) - ( t a n 1 ^{\circ} )}$ , $x > 0$ , then the least value of $f (f (x)) + f (f (\frac{4}{x}))$ is

If $f (x) = lo g_{e} (\frac{1 - x}{1 + x}), ∣ x ∣ < 1$ , then $f (\frac{2 x}{1 + x ^{2}})$ is equal to :

Let $f : R \to R$ be defined as $f (x) = 2 x - 1$ and $g : R - {1} \to R$ be defined as $g (x) = \frac{x - 1}{x}$ . Then the composition function $f (g (x))$ is :

For $x \in R - {0, 1}$ , let $f_{1} (x) = \frac{1}{x}$ , $f_{2} (x) = 1 - x$ and $f_{3} (x) = \frac{1}{1 - x}$ be three given functions. If a function, J(x) satisfies $(f_{2} \circ J \circ f_{1}) (x) = f_{3} (x)$ then J(x) is equal to :-

Let $f (x) = x^{2}$ and $g (x) = sin x$ for all $x \in R$ . Then the set of all $x$ satisfying $(f \circ g \circ g \circ f) (x) = (g \circ g \circ f) (x)$ , where $(f \circ g) (x) = f (g (x))$ , is

Animated Solution for Mathematics - Functions: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x+1αx​,x=−1. Then, for what value of α is f(f(x))=x?