Question

The range of the function $f(x) = \frac{x^2}{x^2 + 1}$ is

• A. $(0, 1)$
• B. $[0, 1)$
• C. $(0, 1]$
• D. $[0, 1]$

Hint:

Algebra Tip: A simpler logical check: since $x^2 \ge 0$, the numerator is positive or zero. The denominator $x^2+1$ is always strictly greater than $x^2$. Thus, a positive number divided by a slightly larger positive number is always less than 1, and the minimum value (when $x=0$) is $0$. Hence, $[0, 1)$.

Answer 1

The Correct Option is B

Concept: Functions - Finding the Range of Rational Functions.

Step 1: Set the function equal to $y$. Let $y = f(x)$. Therefore, $y = \frac{x^{2}}{x^{2}+1}$. To find the range, we need to determine all possible real values that $y$ can take.

Step 2: Express $x^2$ in terms of $y$. Cross-multiply to remove the fraction: $y(x^2 + 1) = x^2 \implies yx^2 + y = x^2$. Rearrange the equation to group all $x^2$ terms on one side: $x^2 - yx^2 = y$. Factor out $x^2$: $x^2(1 - y) = y$. Divide by $(1-y)$ to isolate $x^2$: $x^2 = \frac{y}{1-y}$.

Step 3: Apply the condition for real $x$. For $x$ to be a real number ($x \in \mathbb{R}$), its square must be non-negative. This means $x^2 \ge 0$. Substituting our expression from Step 2, we get the inequality: $\frac{y}{1-y} \ge 0$.

Step 4: Solve the rational inequality. For the fraction $\frac{y}{1-y}$ to be greater than or equal to 0, the numerator and denominator must have the same sign. Also, the denominator cannot be zero, meaning $1-y \neq 0 \implies y \neq 1$.
Let's rewrite the inequality standard form by multiplying the numerator and denominator by -1: $\frac{y}{y-1} \le 0$.
The critical points are $y=0$ (from the numerator) and $y=1$ (from the denominator).
Testing the intervals around these points using the Wavy Curve Method reveals that the inequality is satisfied when $y$ is between 0 and 1.

Step 5: Determine the exact interval bounds. Since the inequality is $\le 0$, we include the numerator's critical point ($y=0$). Thus, $y$ can equal 0. However, $y$ cannot equal 1 because it makes the original denominator zero (and undefined). Therefore, $0 \le y<1$, which is written in interval notation as $[0, 1)$. $$ \therefore \text{The range of the function is } [0, 1). $$