If $f(x) = \dfrac{x - 1}{x + 1}$, then show that (i) $f\left(\dfrac{1}{x}\right) = -f(x)$ (ii) $f\left(-\dfrac{1}{x}\right) = \dfrac{-1}{f(x)}$

> [!TLDR] > The key idea is to substitute the given expression into the function definition and simplify algebraically. For (i), substituting $1/x$ yields $-(x-1)/(x+1) = -f(x)$. For (ii), substituting $-1/x$ gives $(-1)/f(x)$ after simplification. We are given $f(x) = \dfrac{x - 1}{x + 1}$. The function is a rational expression, and the two statements to prove are algebraic identities that hold for all $x$ where both sides are defined (i.e., $x \neq -1$ and $x \neq 0$ for the substitutions). The approach is straightforward: compute $f(1/x)$ and $f(-1/x)$ by replacing $x$ in the formula, then simplify each to match the claimed form. --- ### (i) Show $f\left(\frac{1}{x}\right) = -f(x)$ 1. **Substitute $1/x$ into $f$.** Replace every $x$ in $f(x) = \frac{x-1}{x+1}$ with $1/x$: $$f\left(\frac{1}{x}\right) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1}.$$ 2. **Simplify the numerator and denominator.** Write each as a single fraction over a common denominator $x$: - Numerator: $\frac{1}{x} - 1 = \frac{1 - x}{x}$. - Denominator: $\frac{1}{x} + 1 = \frac{1 + x}{x}$. So $$f\left(\frac{1}{x}\right) = \frac{\frac{1-x}{x}}{\frac{1+x}{x}}.$$ 3. **Cancel the common factor $x$.** Dividing two fractions: $\frac{a/x}{b/x} = \frac{a}{b}$. Hence $$f\left(\frac{1}{x}\right) = \frac{1 - x}{1 + x}.$$ 4. **Factor out a negative sign to match $-f(x)$.** Notice $1 - x = -(x - 1)$. Therefore $$f\left(\frac{1}{x}\right) = \frac{-(x - 1)}{x + 1} = -\frac{x - 1}{x + 1} = -f(x).$$ This completes the proof for part (i). > [!TIP] > A quick check: try $x = 2$. Then $f(2) = (2-1)/(2+1) = 1/3$, and $f(1/2) = (0.5-1)/(0.5+1) = (-0.5)/1.5 = -1/3$, which indeed equals $-f(2)$. The algebra confirms the pattern. --- ### (ii) Show $f\left(-\frac{1}{x}\right) = \frac{-1}{f(x)}$ 1. **Substitute $-1/x$ into $f$.** Replace $x$ with $-1/x$: $$f\left(-\frac{1}{x}\right) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1}.$$ 2. **Simplify numerator and denomin

If f(x) = (x - 1)/(x + 1), then show that (i) f((1 – Class 11 Mathematics

NCERT Exemplar · Q20

f(x) = \dfrac{x - 1}{x + 1}

(i)

f\left(\dfrac{1}{x}\right) = -f(x)

(ii)

f\left(-\dfrac{1}{x}\right) = \dfrac{-1}{f(x)}

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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 key idea is to substitute the given expression into the function definition and simplify algebraically. For (i), substituting $1/x$ yields $-(x-1)/(x+1) = -f(x)$ . For (ii), substituting $-1/x$ gives $(-1)/f(x)$ after simplification.

We are given $f(x) = \dfrac{x - 1}{x + 1}$ . The function is a rational expression, and the two statements to prove are algebraic identities that hold for all $x$ where both sides are defined (i.e., $x \neq -1$ and $x \neq 0$ for the substitutions). The approach is straightforward: compute $f(1/x)$ and $f(-1/x)$ by replacing $x$ in the formula, then simplify each to match the claimed form.

(i) Show $f\left(\frac{1}{x}\right) = -f(x)$

Substitute $1/x$ into $f$ . Replace every $x$ in $f(x) = \frac{x-1}{x+1}$ with $1/x$ :

f\left(\frac{1}{x}\right) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1}.

Simplify the numerator and denominator.

Write each as a single fraction over a common denominator $x$ :
- Numerator: $\frac{1}{x} - 1 = \frac{1 - x}{x}$ .
- Denominator: $\frac{1}{x} + 1 = \frac{1 + x}{x}$ .
So

f\left(\frac{1}{x}\right) = \frac{\frac{1-x}{x}}{\frac{1+x}{x}}.

Cancel the common factor $x$ . Dividing two fractions: $\frac{a/x}{b/x} = \frac{a}{b}$ . Hence

f\left(\frac{1}{x}\right) = \frac{1 - x}{1 + x}.

Factor out a negative sign to match $-f(x)$ . Notice $1 - x = -(x - 1)$ . Therefore

f\left(\frac{1}{x}\right) = \frac{-(x - 1)}{x + 1} = -\frac{x - 1}{x + 1} = -f(x).

This completes the proof for part (i).

Tip

A quick check: try $x = 2$ . Then $f(2) = (2-1)/(2+1) = 1/3$ , and $f(1/2) = (0.5-1)/(0.5+1) = (-0.5)/1.5 = -1/3$ , which indeed equals $-f(2)$ . The algebra confirms the pattern.

(ii) Show $f\left(-\frac{1}{x}\right) = \frac{-1}{f(x)}$

Substitute $-1/x$ into $f$ . Replace $x$ with $-1/x$ :

f\left(-\frac{1}{x}\right) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1}.

Simplify numerator and denominator.

Write each term over $x$ :
- Numerator: $-\frac{1}{x} - 1 = \frac{-1 - x}{x}$ .
- Denominator: $-\frac{1}{x} + 1 = \frac{-1 + x}{x}$ .
So

f\left(-\frac{1}{x}\right) = \frac{\frac{-1 - x}{x}}{\frac{-1 + x}{x}}.

Cancel $x$ and simplify. The $x$ cancels, leaving

f\left(-\frac{1}{x}\right) = \frac{-1 - x}{-1 + x}.

Factor negatives to relate to $f(x)$ . …

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(i) Show f(1x)=−f(x)f\left(\frac{1}{x}\right) = -f(x)f(x1​)=−f(x)

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Q.If $f(x) = \dfrac{x - 1}{x + 1}$ , then show that

(i) Show $f\left(\frac{1}{x}\right) = -f(x)$

(ii) Show $f\left(-\frac{1}{x}\right) = \frac{-1}{f(x)}$

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