Question Statement

Check whether the following functions are one-to-one or not:

(i) $f(x)=4x-7$

(ii) $f(x)=6x^2+2$

(iii) $f(x)=\frac{x^3-1}{2}$

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Background and Explanation

A function $f$ is considered one-to-one (or injective) if every unique input $x$ produces a unique output $f(x)$. To test this algebraically, we assume $f(x_1)=f(x_2)$ and solve to see if it necessarily implies that $x_1=x_2$.

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Solution

Part (i)

Function: $f(x)=4x-7$

To check if the function is one-to-one, we set $f(x_1)=f(x_2)$: $\begin{array}{rl}f\left(x_1\right)& =f\left(x_2\right)\\ 4x_1-7& =4x_2-7\end{array}$ Adding $7$ to both sides: $4x_1=4x_2$ Dividing both sides by $4$: $x_1=x_2$ Since $f(x_1)=f(x_2)$ leads directly to $x_1=x_2$, $f$ is a one-to-one function.

Part (ii)

Function: $f(x)=6x^2+2$

We set $f(x_1)=f(x_2)$: $\begin{array}{rl}6x_1^2+2& =6x_2^2+2\\ 6x_1^2& =6x_2^2\\ x_1^2& =x_2^2\end{array}$ When we take the square root of both sides, we must consider both positive and negative possibilities: $x_1=\pm x_2$ This means $x_1$ does not necessarily equal $x_2$. For example, if we choose $x_1=2$ and $x_2=-2$: $f(2)=6(2{)}^2+2=26$ $f(-2)=6(-2{)}^2+2=26$ Since $f(2)=f(-2)$ but $2\ne -2$, $f$ is not a one-to-one function.

Part (iii)

Function: $f(x)=\frac{x^3-1}{2}$

We set $f(x_1)=f(x_2)$: $\begin{array}{rl}\frac{x_1^3-1}{2}& =\frac{x_2^3-1}{2}\\ x_1^3-1& =x_2^3-1\\ x_1^3& =x_2^3\end{array}$ Taking the cube root of both sides (unlike square roots, cube roots of real numbers preserve the sign and result in a single unique value): $x_1=x_2$ Since $f(x_1)=f(x_2)$ leads to $x_1=x_2$, $f$ is a one-to-one function.

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Key Formulas or Methods Used

• Injective Property: $f(x_1)=f(x_2)⟹x_1=x_2$.
• Algebraic Simplification: Isolating the variable to compare $x_1$ and $x_2$.
• Power Rules:
  • Even powers (like $x^2$) can result in the same value for positive and negative inputs.
  • Odd powers (like $x^3$) maintain a one-to-one relationship for all real numbers.

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Summary of Steps

1. Assume two inputs $x_1$ and $x_2$ produce the same output: $f(x_1)=f(x_2)$.
2. Substitute the inputs into the function expression.
3. Use algebraic operations (addition, subtraction, multiplication, division) to simplify the equation.
4. Analyze the resulting relationship between $x_1$ and $x_2$:
   • If $x_1=x_2$ is the only solution, the function is one-to-one.
   • If $x_1$ can equal something other than $x_2$ (like $-x_2$), the function is not one-to-one.