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

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

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Given: $f(x) = \frac{3}{x} + 7$f(x)=x3​+7

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

Using the power rule:

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

$$f'(x) = -3x^{-2}$$f′(x)=−3x−2

$$f'(x) = -\frac{3}{x^2}$$f′(x)=−x23​

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

$$-\frac{3}{x^2} < 0$$−x23​<0

Since $x^2$x2 is always positive for any $x \neq 0$x=0, the expression $-\frac{3}{x^2}$−x23​ is always negative.

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

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The function $f(x) = \frac{3}{x} + 7$f(x)=x3​+7 is decreasing for $x \in \mathbb{R} - \{0\}$x∈R−{0}.