Let f : N → Y be a function defined as f(x) = 4x + 3 where Y = {y ∈ N : y = 4x + 3 for some x ∈ N}.

Q: Wait, why is the codomain $Y$ defined as $\{y \in N : y = 4x + 3 \text{ for some } x \in N\}$? Can't it just be $N$?

That is a brilliant observation! By defining $Y$ this way, the problem ensures that the codomain is exactly the range. If the codomain were just $N$, the function would not be surjective because many numbers in $N$ (like $1, 2, 4, 5$) are never reached by $4x+3$. This specific definition guarantees surjectivity, which is a mandatory prerequisite for invertibility.

Q: Do I really need to prove injectivity and surjectivity every time?

In a formal JEE Advanced proof, yes! While you might intuitively see that $4x+3$ is a linear, strictly increasing function, the rigorous proof requires showing that $f(x_1) = f(x_2) \implies x_1 = x_2$. It builds the mathematical discipline required for more complex calculus and analysis problems later in your journey.

Q: Can I just swap $x$ and $y$ and solve for $y$?

That is a common shortcut, but be careful! Swapping variables is a technique to find the expression, but it does not prove the function is invertible. Always verify the domain and codomain constraints first. In this problem, the mapping $N \to Y$ is crucial to ensure the inverse is well-defined.

Question

Let $f : N 	o Y$ be a function defined as $f(x) = 4x + 3$ where $Y = \{y \in N : y = 4x + 3 	ext{ for some } x \in N\}$. Show that $f$ is invertible and its inverse is

Options:

(A) $g(y) = \frac{3y + 4}{3}$

(B) $g(y) = 4 + \frac{y + 3}{4}$

(C) $g(y) = \frac{y + 3}{4}$

(D) $g(y) = \frac{y - 3}{4}$

Sigma Percentile · Accepted Answer

### The Architecture of Reversibility: Unlocking the Inverse Function

Imagine you are standing before a complex machine. You feed it a number, $x$, and it performs a specific operation: it multiplies your input by $4$ and adds $3$.

The machine spits out a result, $y$. Now, imagine you are holding that result, $y$, and you want to know exactly what number you fed into the machine to get it. This is the essence of an inverse function; it is the machine running in reverse.

### The Bijective Requirement

Before we can reverse the machine, we must ensure it is a 'bijective' machine. Think of this as the 'no-collision' rule.

If two different inputs, $x_1$ and $x_2$, produced the same output, $y$, then when we tried to reverse the machine, we would be confused—we wouldn't know which input to return to! This is why we must prove injectivity.

We assume $f(x_1) = f(x_2)$, which leads us to the following equation:

$$4x_1 + 3 = 4x_2 + 3$$

By subtracting $3$ from both sides and dividing by $4$, we arrive at $x_1 = x_2$. This confirms that our machine is perfectly injective; every input has a unique, distinct output.

### The Surjective Guarantee

Next, we must ensure that every possible output in our destination set $Y$ is actually reachable. If there were a number in $Y$ that the machine could never produce, the inverse function would have nowhere to send that number.

The problem defines $Y$ as the set of all $y$ such that $y = 4x + 3$. By definition, the codomain is the range. This is the 'onto' or surjective property.

Because our function is both injective and surjective, it is a bijection, and thus, it is invertible.

### The Reverse Engineering

Now, for the thrill of the solution. We have the equation $y = 4x + 3$. Our goal is to isolate $x$ to find the 'reverse' rule.

We subtract $3$ from both sides to get $y - 3 = 4x$. Then, we divide by $4$ to isolate $x$, yielding:

$$x = \frac{y - 3}{4}$$

This expression is our inverse function, $g(y) = \frac{y - 3}{4}$. It is elegant, precise, and perfectly reverses the original operation.

You have successfully navigated the logic of functions, proving that with the right mathematical tools, any transformation can be undone. Keep this clarity of thought as you approach more complex problems; the logic remains the same, even when the functions become more intricate.

Animated Solution for Mathematics - Functions: Let $f : N \to Y$ be a function defined as $f (x) = 4 x + 3$ where $Y = {y \in N : y = 4 x + 3 for some x \in N}$ . Show that $f$ is invertible and its inverse is

Select Answer:

Visualized Solution

Function Definition $f : N \to Y$

Condition for Invertibility

Proving Injectivity ( $1 - 1$ )

Substituting the Function Rule

Algebraic Simplification

Confirming One-to-one

Proving Surjectivity (Onto)

Setting up the Inverse

Transposing the Constant

Isolating $x$

Final Result: The Inverse Function

The Sigma Insight: Inverse of a Function

The Architecture of Reversibility

The Bijective Requirement

The Surjective Guarantee

The Reverse Engineering

Similar Questions

Let $f : A \to B$ be a function defined as $f (x) = \frac{x - 1}{x - 2}$ , where $A = R - {2}$ and $B = R - {1}$ . Then f is :-

If $f : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + 1/ x$ , then $f^{- 1} (x)$ equals

If $f (x) = sin x + cos x, g (x) = x^{2} - 1$ , then $g (f (x))$ is invertible in the domain

Let $f$ be a one-one function with domain ${x, y, z}$ and range ${1, 2, 3}$ . It is given that exactly one of the following statements is true and the remaining two are false $f (x) = 1, f (y) \neq = 1, f (z) \neq = 2$ determine $f^{- 1} (1)$ .

The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}$ , $x \in (- 1, 1)$ , is

The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}$ , $x \in (- 1, 1)$ , is

If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ , then $f^{- 1} (x)$ is

Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x}$ , where $b$ is a constant such that $0 < b < 1$ . Then

If $f (x) = 3 x - 5$ , then $f^{- 1} (x)$

Let $f$ be a real-valued function defined on the interval (-1, 1) such that $e^{- x} f (x) = 2 + \int_{0}^{x} t^{4} + 1 d t$ , for all $x \in (- 1, 1)$ , and let $f^{- 1}$ be the inverse function of $f$ . Then $(f^{- 1})^{'} (2)$ is equal to

Select Answer:

Visualized Solution

Function Definition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f:N→Y

Condition for Invertibility

Proving Injectivity ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1−1)

Substituting the Function Rule

Algebraic Simplification

Confirming One-to-one

Proving Surjectivity (Onto)

Setting up the Inverse

Transposing the Constant

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Final Result: The Inverse Function

The Sigma Insight: Inverse of a Function

The Architecture of Reversibility

The Bijective Requirement

The Surjective Guarantee

The Reverse Engineering

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:A→B be a function defined as f(x)=x−2x−1​, where A=R−{2} and B=R−{1}. Then f is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f:[1,∞)→[2,∞) is given by f(x)=x+1/x, then f−1(x) equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=sinx+cosx,g(x)=x2−1, then g(f(x)) is invertible in the domain

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The inverse function of f(x)=82x+8−2x82x−8−2x​, x∈(−1,1), is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The inverse function of f(x)=82x+8−2x82x−8−2x​, x∈(−1,1), is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the function f:[1,∞)→[1,∞) is defined by f(x)=2x(x−1), then f−1(x) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:(0,1)→R be defined by f(x)=1−bxb−x​, where b is a constant such that 0<b<1. Then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=3x−5, then f−1(x)

Function Definition $f : N \to Y$

Proving Injectivity ( $1 - 1$ )

Isolating $x$

Let $f : A \to B$ be a function defined as $f (x) = \frac{x - 1}{x - 2}$ , where $A = R - {2}$ and $B = R - {1}$ . Then f is :-

If $f : [1, \infty) \to [2, \infty)$ is given by $f (x) = x + 1/ x$ , then $f^{- 1} (x)$ equals

If $f (x) = sin x + cos x, g (x) = x^{2} - 1$ , then $g (f (x))$ is invertible in the domain

The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}$ , $x \in (- 1, 1)$ , is

The inverse function of $f (x) = \frac{8 ^{2 x} - 8 ^{- 2 x}}{8 ^{2 x} + 8 ^{- 2 x}}$ , $x \in (- 1, 1)$ , is

If the function $f : [1, \infty) \to [1, \infty)$ is defined by $f (x) = 2^{x (x - 1)}$ , then $f^{- 1} (x)$ is

Let $f : (0, 1) \to R$ be defined by $f (x) = \frac{b - x}{1 - b x}$ , where $b$ is a constant such that $0 < b < 1$ . Then

If $f (x) = 3 x - 5$ , then $f^{- 1} (x)$