If $f(x) = y = \dfrac{ax - b}{cx - a}$, then prove that $f(y) = x$.

> [!TLDR] > The function $f(x) = \frac{ax - b}{cx - a}$ is an **involution** — applying it twice returns the original input. Substituting $y$ into $f$ and simplifying using the relation $y = f(x)$ gives $f(y) = x$. The core idea here is that the function has a special symmetry: it is its own inverse. When you see a rational function of the form $\frac{ax - b}{cx - a}$, the constants are arranged so that the numerator and denominator swap roles in a neat way. Instead of finding the inverse function explicitly, we can directly substitute $y$ into $f$ and simplify using the given relation $y = f(x)$. This avoids solving for $x$ in terms of $y$ and is much cleaner. Let’s work through it step by step. 1. **Write down what we know.** We are given $f(x) = y = \frac{ax - b}{cx - a}$. This means $y$ is expressed in terms of $x$. We want to prove $f(y) = x$, i.e., $f(y) = \frac{a y - b}{c y - a}$ simplifies to $x$. 2. **Substitute $y$ into $f$.** $$f(y) = \frac{a y - b}{c y - a}.$$ Replace $y$ with $\frac{ax - b}{cx - a}$: $$f(y) = \frac{a\left(\frac{ax - b}{cx - a}\right) - b}{c\left(\frac{ax - b}{cx - a}\right) - a}.$$ 3. **Simplify the numerator.** Combine the terms in the numerator over a common denominator: $$a\left(\frac{ax - b}{cx - a}\right) - b = \frac{a(ax - b)}{cx - a} - b = \frac{a(ax - b) - b(cx - a)}{cx - a}.$$ Expand: $$a(ax - b) = a^2 x - ab, \quad b(cx - a) = bcx - ab.$$ So the numerator becomes: $$\frac{(a^2 x - ab) - (bcx - ab)}{cx - a} = \frac{a^2 x - ab - bcx + ab}{cx - a} = \frac{a^2 x - bcx}{cx - a}.$$ Factor $x$: $$\frac{x(a^2 - bc)}{cx - a}.$$ 4. **Simplify the denominator.** Similarly, the denominator of $f(y)$ is: $$c\left(\frac{ax - b}{cx - a}\right) - a = \frac{c(ax - b)}{cx - a} - a = \frac{c(ax - b) - a(cx - a)}{cx - a}.$$ Expand: $$c(ax - b) = acx - bc, \quad a(cx - a) = acx - a^2.$$ So the denominator becomes: …

If f(x) = y = (ax - b)/(cx - a), then prove that f – Class 11 Mathematics

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f(x) = y = \dfrac{ax - b}{cx - a}

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Concept understanding — Function Evaluation

Function Evaluation: What It Really Means

Imagine you have a machine. You drop a number into the top, the machine does something to it, and a new number comes out the bottom. That machine is a function. Function evaluation is simply the act of putting a specific input into that machine and seeing what output you get.

Let’s make it concrete. Suppose the machine’s rule is: “Take whatever number you get, multiply it by 2, then add 3.” If you drop in a 5, the machine does $2 \times 5 + 3 = 13$ , and out comes 13. You have just evaluated the function at the input 5.

Note

The input is often called the argument or the independent variable. The output is the value of the function at that argument.

The Precise Statement

A function is a rule that assigns to every input (from a set called the domain) exactly one output (from a set called the codomain). We usually name a function with a letter like $f$ , $g$ , or $h$ . The input is often written as $x$ , and the output as $f(x)$ (read “f of x”).

So if $f$ is the function “multiply by 2 and add 3,” we write:

f(x) = 2x + 3

To evaluate $f$ at $x = 5$ , we replace every $x$ in the expression with 5:

f(5) = 2(5) + 3 = 10 + 3 = 13

That’s it. Function evaluation is substitution: take the given input, plug it into the function’s rule wherever the variable appears, and simplify.

Why This Matters

Every formula you’ve ever seen — area of a circle $A(r) = \pi r^2$ , distance $d(t) = vt$ , even the quadratic formula — is a function. Evaluating it means you’re answering a specific question: “What happens when the input is this particular number?”

Tip

When you see $f(2)$ , read it as “the output when the input is 2.” Don’t think of it as multiplication — $f$ is not multiplied by 2. It’s a notation for the result of the rule.

A Common Mistake

Students sometimes write $f(x+1) = 2x+3$ and then try to “solve” for $x$ . Don’t. If $f(x) = 2x+3$ , then $f(x+1)$ means: replace every $x$ in the rule with $(x+1)$ :

f(x+1) = 2(x+1) + 3 = 2x + 2 + 3 = 2x + 5

The input is the whole expression $x+1$ , not just $x$ .

The Big Picture …

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The function $f(x) = \frac{ax - b}{cx - a}$ is an involution — applying it twice returns the original input. Substituting $y$ into $f$ and simplifying using the relation $y = f(x)$ gives $f(y) = x$ .

The core idea here is that the function has a special symmetry: it is its own inverse. When you see a rational function of the form $\frac{ax - b}{cx - a}$ , the constants are arranged so that the numerator and denominator swap roles in a neat way. Instead of finding the inverse function explicitly, we can directly substitute $y$ into $f$ and simplify using the given relation $y = f(x)$ . This avoids solving for $x$ in terms of $y$ and is much cleaner.

Let’s work through it step by step.

Write down what we know.

We are given $f(x) = y = \frac{ax - b}{cx - a}$ . This means $y$ is expressed in terms of $x$ . We want to prove $f(y) = x$ , i.e., $f(y) = \frac{a y - b}{c y - a}$ simplifies to $x$ .
Substitute $y$ into $f$ .

f(y) = \frac{a y - b}{c y - a}.

Replace $y$ with $\frac{ax - b}{cx - a}$ :

f(y) = \frac{a\left(\frac{ax - b}{cx - a}\right) - b}{c\left(\frac{ax - b}{cx - a}\right) - a}.

Simplify the numerator. Combine the terms in the numerator over a common denominator:

a\left(\frac{ax - b}{cx - a}\right) - b = \frac{a(ax - b)}{cx - a} - b = \frac{a(ax - b) - b(cx - a)}{cx - a}.

Expand:

a(ax - b) = a^2 x - ab, \quad b(cx - a) = bcx - ab.

So the numerator becomes:

\frac{(a^2 x - ab) - (bcx - ab)}{cx - a} = \frac{a^2 x - ab - bcx + ab}{cx - a} = \frac{a^2 x - bcx}{cx - a}.

Factor $x$ :

\frac{x(a^2 - bc)}{cx - a}.

Simplify the denominator. Similarly, the denominator of $f(y)$ is:

c\left(\frac{ax - b}{cx - a}\right) - a = \frac{c(ax - b)}{cx - a} - a = \frac{c(ax - b) - a(cx - a)}{cx - a}.

Expand:

c(ax - b) = acx - bc, \quad a(cx - a) = acx - a^2.

So the denominator becomes: …

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Q.If f(x)=y=ax−bcx−af(x) = y = \dfrac{ax - b}{cx - a}f(x)=y=cx−aax−b​, then prove that f(y)=xf(y) = xf(y)=x.

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Q.If $f(x) = y = \dfrac{ax - b}{cx - a}$ , then prove that $f(y) = x$ .

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