The largest open interval, in which the function f(x) = x/(x^2 + 1) increases, is Option 1: (0, 1) Option 2: (-1, 0) Option 3: (-1, 1) Option 4: (-∞, -1) (1, ∞)

Answer: (-1, 1). Solution: A function increases where its derivative is positive, i.e., f'(x) > 0. The function is f(x) = x/(x^2 + 1) Using the quotient rule where u = x and v = x^2 + 1: f'(x) = ((1)(x^2 + 1) - (x)(2x))/((x^2 + 1)^2) f'(x) = (x^2 + 1 - 2x^2)/((x^2 + 1)^2) f'(x) = (1 - x^2)/((x^2 + 1)^2) For the function to increase: (1 - x^2)/((x^2 + 1)^2) > 0 The denominator (x^2 + 1)^2 is always positive since x^2 + 1 ≥ 1 for all real x. Therefore, the sign of f'(x) depends only on the numerator: 1 - x^2 > 0 1 > x^2 x^2 The inequality x^2 < 1 holds when: -1 < x < 1 Therefore, the function increases in the interval (-1, 1).

Calculus

Application of Derivatives

Medium

core

The largest open interval, in which the function $f(x) = \frac{x}{x^2 + 1}$ increases, is

$(0, 1)$

$(-1, 0)$

$(-1, 1)$

$(-\infty, -1) \cup (1, \infty)$

✅ Correct Option: 3

A function increases where its derivative is positive, i.e., $f'(x) > 0$ .

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

Using the quotient rule where $u = x$ and $v = x^2 + 1$ :

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

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

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

For the function to increase:

$\frac{1 - x^2}{(x^2 + 1)^2} > 0$

The denominator $(x^2 + 1)^2$ is always positive since $x^2 + 1 \geq 1$ for all real $x$ .

Therefore, the sign of $f'(x)$ depends only on the numerator:

$1 - x^2 > 0$

$1 > x^2$

$x^2 < 1$

The inequality $x^2 < 1$ holds when:

$-1 < x < 1$

Therefore, the function increases in the interval $(-1, 1)$ .

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The largest open interval, in which the function $f(x) = \frac{x}{x^2 + 1}$ increases, is