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The problem asks to calculate \( f\left(-\frac{1}{5}\right) \), given that:

\[
f(x) = \frac{2x + 3}{x^2} + \frac{1}{5x^3}
\]

To solve this, we substitute \( x = -\frac{1}{5} \) into the function \( f(x) \).

### Step 1: Substitute \( x = -\frac{1}{5} \) into the equation
The function becomes:

\[
f\left(-\frac{1}{5}\right) = \frac{2\left(-\frac{1}{5}\right) + 3}{\left(-\frac{1}{5}\right)^2} + \frac{1}{5\left(-\frac{1}{5}\right)^3}
\]

### Step 2: Simplify each part

1. **First term**: 
   \[
   \frac{2\left(-\frac{1}{5}\right) + 3}{\left(-\frac{1}{5}\right)^2} = \frac{-\frac{2}{5} + 3}{\left(\frac{1}{25}\right)} = \frac{\frac{13}{5}}{\frac{1}{25}} = 13 \times 5 = 65
   \]

2. **Second term**: 
   \[
   \frac{1}{5\left(-\frac{1}{5}\right)^3} = \frac{1}{5\left(-\frac{1}{125}\right)} = \frac{1}{-\frac{1}{25}} = -25
   \]

### Step 3: Add the results
\[
f\left(-\frac{1}{5}\right) = 65 + (-25) = 40
\]

Thus, the value of \( f\left(-\frac{1}{5}\right) \) is 40.

Would you like more details or have any questions?

Here are five related questions:
1. How do you find the inverse of a cubic term?
2. What are common mistakes when dealing with fractional exponents?
3. How does the sign change in cube functions affect the outcome?
4. How do you differentiate between reciprocal functions and direct proportionality?
5. How can fractional powers be simplified in general terms?

**Tip:** Always remember to simplify powers and negative signs before combining terms in functions involving fractions.

Calculate f(-1/5) if f(x) = (2x + 3)/x^2 + 1/(5x^3).

This problem involves calculating f(-1/5) for the function f(x) = (2x + 3)/x^2 + 1/(5x^3). It covers important algebraic concepts like substitution and simplifying fractional expressions. This is suitable for students in Grades 10-12.

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