If f : [0, ∞) → [0, ∞), and f(x) = x / (1 + x) then f is (a) one-one and onto (b) one-one but not on

Q: Wait, why do we check the derivative to see if it's one-one?

Great question! A function is one-one (injective) if it never takes the same value twice. If a function is strictly increasing or strictly decreasing, it will always be moving in one direction, meaning it can never return to a previous $y$-value. By showing $f'(x) > 0$, we prove the function is strictly increasing, which guarantees it is one-one.

Q: Can I just plug in values to check if it's onto?

Plugging in values can give you a hint, but it's not a rigorous proof. To be onto (surjective), the range must cover the *entire* codomain. For this function, we need to find the range by looking at the minimum value at $x=0$ and the horizontal asymptote as $x \to \infty$. This tells us the range is $[0, 1)$, which is clearly not the same as $[0, \infty)$.

Q: Why is the range $[0, 1)$ and not $[0, 1]$?

Excellent observation! The function $f(x) = \frac{x}{1+x}$ approaches $1$ as $x$ goes to infinity, but it never actually reaches $1$. If you try to solve $\frac{x}{1+x} = 1$, you get $x = 1+x$, which implies $0=1$, a contradiction. Thus, $1$ is not in the range, making it an open interval at $1$.

Question

If $f : [0, \infty) 	o [0, \infty)$, and $f(x) = \frac{x}{1 + x}$ then $f$ is

Options:

(A) one-one and onto

(B) one-one but not onto

(C) onto but not one-one

(D) neither one-one nor onto

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a mathematical machine. You feed it a number $x$ from the domain $[0, \infty)$, and it spits out a value defined by the function:

$$f(x) = \frac{x}{1+x}$$

Our goal is to understand the nature of this machine: is it one-one? Is it onto? Let's embark on this journey.

### The One-One Investigation

To determine if a function is one-one, we need to know if it ever repeats itself. If the machine is always moving in one direction—either strictly increasing or strictly decreasing—it will never return to a previous output.

This is where the power of calculus shines. We calculate the derivative $f'(x)$ using the quotient rule:

$$f'(x) = \frac{(1+x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(1+x)}{(1+x)^2}$$

Simplifying this expression, we obtain:

$$f'(x) = \frac{1+x-x}{(1+x)^2} = \frac{1}{(1+x)^2}$$

Since $(1+x)^2$ is always positive for $x \in [0, \infty)$, our derivative $f'(x)$ is strictly greater than zero. This tells us the function is **strictly increasing**.

Because it is always climbing, it can never hit the same $y$-value twice. Thus, we have rigorously proven that the function is **one-one**.

### The Onto Investigation

Now, we turn our attention to the second property: surjectivity, or being "onto." A function is onto if its range—the set of all outputs it actually produces—perfectly matches its codomain.

We are given the codomain $[0, \infty)$. Let's find the range. Since the function is strictly increasing, its minimum value occurs at the start of the domain, $x=0$.

Plugging this in, we find:

$$f(0) = \frac{0}{1+0} = 0$$

So, the range starts at $0$. But what is the upper limit? We look at the behavior as $x$ grows infinitely large:

$$\lim_{x 	o \infty} \frac{x}{1+x} = \lim_{x 	o \infty} \frac{1}{\frac{1}{x} + 1} = 1$$

The function approaches $1$ but never reaches it. Therefore, the range is the interval $[0, 1)$.

### The Final Verdict

We have discovered that the range is $[0, 1)$, while the codomain is $[0, \infty)$. Since the range is a proper subset of the codomain, the function **fails to be onto**.

It covers only a small slice of the target space. We have successfully navigated the problem: the function is **one-one** because it is strictly increasing, but it is **not onto** because its range is restricted.

It is a beautiful example of how calculus allows us to map the behavior of functions with absolute precision.

Animated Solution for Mathematics - Functions: If $f : [0, \infty) \to [0, \infty)$ , and $f (x) = \frac{x}{1 + x}$ then $f$ is

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Visualized Solution

Function Definition and Domain

Checking One-One: The Derivative Approach

Applying the Quotient Rule

Simplifying the Derivative

Conclusion on One-One (Injectivity)

Checking Onto: The Range Approach

Finding the Range: Lower Bound

Finding the Range: Upper Bound

Conclusion on Onto (Surjectivity)

Final Answer

The Sigma Insight: Classification of Functions

Analyzing the Setup

The One-One Investigation

The Onto Investigation

The Final Verdict

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