1,008,000/Q + 6.5Q find the instantaneous rate of change when Q=350

To find the instantaneous rate of change for the given function \( f(Q) = \frac{1,008,000}{Q} + 6.5Q \) when \( Q = 350 \), we need to follow these steps:

### 1. Differentiate the function with respect to \( Q \).
We need to find \( f'(Q) \), the derivative of \( f(Q) \).

\[
f(Q) = \frac{1,008,000}{Q} + 6.5Q
\]

The derivative of \( f(Q) \) is:

\[
f'(Q) = -\frac{1,008,000}{Q^2} + 6.5
\]

### 2. Plug in \( Q = 350 \) into \( f'(Q) \).

\[
f'(350) = -\frac{1,008,000}{350^2} + 6.5
\]

Let’s compute that.

\[
350^2 = 122,500
\]

\[
f'(350) = -\frac{1,008,000}{122,500} + 6.5 = -8.23 + 6.5 = -1.73
\]

### 3. Conclusion:
The instantaneous rate of change of the function when \( Q = 350 \) is approximately **-1.73**.

Would you like more details on any step?

Here are 5 related questions to expand your understanding:
1. What is the meaning of an instantaneous rate of change in real-life situations?
2. How do you interpret the result of an instantaneous rate of change being negative?
3. How does this rate of change relate to velocity in physics problems?
4. Can you explain how to differentiate functions with both fractions and variables?
5. What does the second derivative of this function represent?

**Tip**: The instantaneous rate of change of a function at a given point is essentially the slope of the tangent line to the curve at that point.

This solution explains how to find the instantaneous rate of change for the function f(Q) = 1,008,000/Q + 6.5Q when Q=350 using differentiation. Learn the key steps and formulas used to calculate the result of -1.73. Suitable for advanced high school or college students studying calculus and algebra.

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