Q.Prove that the function $f : \mathbf{R} \to \mathbf{R}$ defined by $f(x) = 2x + 5$ is one-one.

$f$ is one-one if $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$. Assume $f(x_1) = f(x_2)$:

Thus the equal outputs force equal inputs.

We use the direct method. Recall the definition: a function $f$ is one-one (injective) if

Step 1 — Assume equal outputs.

Let $x_1, x_2 \in \mathbf{R}$ be such that $f(x_1) = f(x_2)$. By the definition of $f$,

Step 2 — Subtract $5$ from both sides.

Step 3 — Divide both sides by the non-zero number $2$.

Starting from $f(x_1) = f(x_2)$ we have deduced $x_1 = x_2$, which is exactly the condition for $f$ to be one-one.

Method: Proving a Function is One-One by the Direct (Algebraic) Method

This method applies to any "prove $f$ is one-one" question for a function given by an explicit formula.

Steps

Step 1: State the definition you will use

Recall that $f$ is one-one if $f(x_1) = f(x_2) \implies x_1 = x_2$ for all $x_1, x_2$ in the domain — this contrapositive form ("equal outputs force equal inputs") is the version you actually prove, since it turns into a chain of algebra.

Step 2: Assume the outputs are equal, for two arbitrary inputs

Let $x_1, x_2$ be arbitrary elements of the domain with $f(x_1) = f(x_2)$. This must be done for general $x_1, x_2$ — never for two specific numbers, since that would only check one pair, not prove the property for every pair.

Step 3: Substitute the formula for $f$ and simplify

Write out $f(x_1) = f(x_2)$ using the actual rule defining $f$, then use ordinary algebra (adding/subtracting the same quantity from both sides, dividing by a nonzero constant) to isolate $x_1$ and $x_2$.

Step 4: Conclude $x_1 = x_2$

Once the algebra reduces to $x_1 = x_2$, state explicitly that this is exactly the definition of one-one, so $f$ is one-one.

Applying to this problem: with $f(x) = 2x+5$, assume $2x_1+5 = 2x_2+5$, subtract $5$ and divide by the nonzero constant $2$ to get $x_1 = x_2$.

Common Mistakes

Mistake 1: Testing with specific numbers instead of general $x_1, x_2$

Why it's wrong: Checking that, say, $f(1) \ne f(2)$ only verifies one pair of inputs — it says nothing about every other pair, so it can never establish that $f$ is one-one for all $x_1, x_2$. Correct approach: let $x_1$ and $x_2$ be arbitrary (unspecified) real numbers and show the implication holds symbolically.

Mistake 2: Treating the horizontal-line test as a substitute for the algebraic proof

Why it's wrong: The horizontal-line test is a useful graphical check, but on its own it is not a rigorous proof — the question asks to "prove" $f$ is one-one, which requires the algebraic $f(x_1)=f(x_2) \Rightarrow x_1=x_2$ argument, not just a visual justification. Correct approach: use the algebraic direct method as the actual proof, and treat any graphical reasoning as intuition only.