The function `f(x) = 3/x + 10, x ne 0` is decreasing

Answer

To determine if the function \( f(x) = \frac{3}{x} + 10 \) (where \( x \neq 0 \)) is decreasing, we need to analyze its derivative.

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = \frac{3}{x} + 10 \]

2. **Differentiate the function**:
To find if the function is decreasing, we need to calculate the derivative \( f'(x) \).
\[ f'(x) = \frac{d}{dx}\left(\frac{3}{x}\right) + \frac{d}{dx}(10) \]
The derivative of \( \frac{3}{x} \) can be calculated using the power rule:
\[ \frac{3}{x} = 3x^{-1} \implies \frac{d}{dx}(3x^{-1}) = -3x^{-2} = -\frac{3}{x^2} \]
The derivative of a constant (10) is 0. Therefore:
\[ f'(x) = -\frac{3}{x^2} \]

3. **Analyze the sign of the derivative**:
The expression \( -\frac{3}{x^2} \) is negative for all \( x \neq 0 \) because \( x^2 \) is always positive for any real number \( x \) (except zero, which is excluded in our function). Thus:
\[ f'(x) < 0 \quad \text{for all } x \neq 0 \]

4. **Conclusion**:
Since \( f'(x) < 0 \), this implies that the function \( f(x) \) is decreasing for all \( x \neq 0 \). Therefore, the statement that the function is decreasing is **true**.

### Final Answer:
The function \( f(x) = \frac{3}{x} + 10 \) is decreasing for \( x \neq 0 \). Thus, the statement is **true**.

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