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The problem asks to show that the function \( f(x) = \frac{3 - 4x}{e^{2x}} \) has a minimum value when \( x = \frac{5}{4} \).

### Steps to verify this:

1. **Function and Critical Points**:
   To find the critical points, we first differentiate the function \( f(x) \).

   The function is:
   \[
   f(x) = \frac{3 - 4x}{e^{2x}}
   \]
   
   Using the quotient rule for differentiation:
   \[
   f'(x) = \frac{(e^{2x}) \cdot \frac{d}{dx}(3 - 4x) - (3 - 4x) \cdot \frac{d}{dx}(e^{2x})}{(e^{2x})^2}
   \]
   Simplifying:
   \[
   f'(x) = \frac{-4e^{2x} - (3 - 4x) \cdot 2e^{2x}}{e^{4x}}
   \]
   \[
   f'(x) = \frac{-4e^{2x} - 2(3 - 4x)e^{2x}}{e^{4x}}
   \]
   Factor out \( e^{2x} \) from the numerator:
   \[
   f'(x) = \frac{e^{2x}[-4 - 2(3 - 4x)]}{e^{4x}} = \frac{-4 - 6 + 8x}{e^{2x}} = \frac{8x - 10}{e^{2x}}
   \]
   
2. **Set the derivative equal to zero** to find the critical points:
   \[
   8x - 10 = 0
   \]
   \[
   x = \frac{10}{8} = \frac{5}{4}
   \]
   
   Therefore, the critical point is \( x = \frac{5}{4} \).

3. **Second Derivative Test**:
   To confirm that this is a minimum, we can check the second derivative \( f''(x) \). Differentiating \( f'(x) \):
   \[
   f'(x) = \frac{8x - 10}{e^{2x}}
   \]
   Differentiate again using the quotient rule:
   \[
   f''(x) = \frac{8e^{2x} - (8x - 10) \cdot 2e^{2x}}{e^{4x}} = \frac{8e^{2x} - 2(8x - 10)e^{2x}}{e^{4x}}
   \]
   Simplifying:
   \[
   f''(x) = \frac{e^{2x}[8 - 2(8x - 10)]}{e^{4x}} = \frac{8 - 16x + 20}{e^{2x}} = \frac{28 - 16x}{e^{2x}}
   \]
   At \( x = \frac{5}{4} \):
   \[
   f''\left(\frac{5}{4}\right) = \frac{28 - 16 \cdot \frac{5}{4}}{e^{2 \cdot \frac{5}{4}}} = \frac{28 - 20}{e^{5/2}} = \frac{8}{e^{5/2}} > 0
   \]
   
   Since \( f''\left(\frac{5}{4}\right) > 0 \), the function has a local minimum at \( x = \frac{5}{4} \).

Thus, the function \( f(x) \) has a minimum value at \( x = \frac{5}{4} \), which matches the condition given in the problem.

Would you like a more detailed explanation or any clarifications on this? 

Here are 5 related questions you might find helpful:
1. How do we use the quotient rule in differentiation?
2. What does the second derivative test indicate about a function's concavity?
3. Can a function have more than one critical point?
4. How is the second derivative related to the shape of the graph?
5. What are other methods to find critical points of a function?

**Tip:** Always check the second derivative when you want to confirm whether a critical point is a minimum or maximum!

Illustrate that the graph of f(x) = (3 - 4x) / e^(2x) has minimum value at x = 5/4.

This calculus problem illustrates how to determine the minimum value of the function f(x) = (3 - 4x) / e^(2x) using differentiation and the second derivative test. Detailed steps include applying the quotient rule and confirming the result at x = 5/4.

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