For x ∈ ℝ - 0, the function f(x) = 3/x + 7 is decreasing when Option 1: x ∈ ℝ Option 2: x ∈ ℝ - 0 Option 3: x = 0 Option 4: x ∈ ∅; ∅ is empty set

Answer: x ∈ ℝ - 0. Solution: A function is decreasing when its derivative is negative, i.e., when f'(x) Given: f(x) = 3/x + 7 Rewriting: f(x) = 3x^(-1) + 7 Using the power rule: f'(x) = 3 * (-1) * x^(-2) + 0 f'(x) = -3x^(-2) f'(x) = -3/x^2 For the function to be decreasing: f'(x) The function f(x) = 3/x + 7 is decreasing for x R - \0\.

Calculus

Application of Derivatives

Hard

core

For x ∈ ℝ - {0}, the function f(x) = $\frac{3}{x}$ + 7 is decreasing when

x ∈ ℝ

x ∈ ℝ - {0}

x = {0}

x ∈ ∅; ∅ is empty set

✅ Correct Option: 2

A function is decreasing when its derivative is negative, i.e., when $f'(x) < 0$ .

Given: $f(x) = \frac{3}{x} + 7$

Rewriting: $f(x) = 3x^{-1} + 7$

Using the power rule:

$f'(x) = 3 \cdot (-1) \cdot x^{-2} + 0$

$f'(x) = -3x^{-2}$

$f'(x) = -\frac{3}{x^2}$

For the function to be decreasing: $f'(x) < 0$

$-\frac{3}{x^2} < 0$

Since $x^2$ is always positive for any $x \neq 0$ , the expression $-\frac{3}{x^2}$ is always negative.

Therefore, $f'(x) < 0$ for all $x \in \mathbb{R} - \{0\}$ .

The function $f(x) = \frac{3}{x} + 7$ is decreasing for $x \in \mathbb{R} - \{0\}$ .

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For x ∈ ℝ - {0}, the function f(x) = 3x\frac{3}{x}x3​ + 7 is decreasing when

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For x ∈ ℝ - {0}, the function f(x) = $\frac{3}{x}$ + 7 is decreasing when