The function, f(x) = x - 1/x is Option 1: increasing for all x (-∞, 0) (0, ∞) Option 2: decreasing for all x (-∞, 0) (0, ∞) Option 3: increasing for all x (-∞, ∞) Option 4: neither increasing nor decreasing for x (-∞, ∞)

Answer: increasing for all x (-∞, 0) (0, ∞). Solution: To determine whether f(x) = x - 1/x is increasing or decreasing, find the derivative and check its sign. f(x) = x - 1/x Rewrite as: f(x) = x - x^(-1) Differentiate: f'(x) = 1 - (-1)x^(-2) f'(x) = 1 + 1/x^2 For f'(x) = 1 + 1/x^2: Since x^2 is always positive for x ≠ 0, then 1/x^2 is always positive. Therefore: f'(x) = 1 + 1/x^2 > 1 > 0 Since f'(x) > 0 for all x ≠ 0, the function is increasing. The function f(x) = x - 1/x is not defined at x = 0 (division by zero). The domain is x (-∞, 0) (0, ∞). Therefore, the function is increasing for all x (-∞, 0) (0, ∞).

Calculus

Application of Derivatives

Medium

common

The function, $f(x) = x - \frac{1}{x}$ is

increasing for all $x \in (-\infty, 0) \cup (0, \infty)$

decreasing for all $x \in (-\infty, 0) \cup (0, \infty)$

increasing for all $x \in (-\infty, \infty)$

neither increasing nor decreasing for $x \in (-\infty, \infty)$

✅ Correct Option: 1

To determine whether $f(x) = x - \frac{1}{x}$ is increasing or decreasing, find the derivative and check its sign.

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

Rewrite as:

$f(x) = x - x^{-1}$

Differentiate:

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

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

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

Since $x^2$ is always positive for $x \neq 0$ , then $\frac{1}{x^2}$ is always positive.

Therefore:

$f'(x) = 1 + \frac{1}{x^2} > 1 > 0$

Since $f'(x) > 0$ for all $x \neq 0$ , the function is increasing.

The function $f(x) = x - \frac{1}{x}$ is not defined at $x = 0$ (division by zero).

The domain is $x \in (-\infty, 0) \cup (0, \infty)$ .

Therefore, the function is increasing for all $x \in (-\infty, 0) \cup (0, \infty)$ .

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Match list I with list II

List - I	List - II
A. Slope of tangent to the curve $y = x^3 - x$ at $x = 2$	I. $-2$
B. Slope of tangent to the curve $y = 3x^3 - 4x$ at $x = 0$	II. $11$
C. Slope of normal to the curve $y = \sin\theta$ at $\theta = \frac{\pi}{3}$	III. $2$
D. Slope of normal to the curve $y = \cos\theta$ at $\theta = \frac{\pi}{6}$	IV. $-4$

Choose the correct answer from the option given below :

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The function $f(x) = x^2 - 4x + 6$ is

(A) Strictly decreasing on $(-\infty, 2) \cup (2, \infty)$

(B) Strictly increasing on $(2, \infty)$

(D) Strictly decreasing on $(-\infty, 2)$

Choose the correct answer from the options given below:

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The rate of change of area of a circle with respect to its circumference when radius in 6 cm, is

The function, f(x)=x−1xf(x) = x - \frac{1}{x}f(x)=x−x1​ is

Solution

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The function, $f(x) = x - \frac{1}{x}$ is