To find the value of $a$, we start by using the information given in the problem.

The equation of the curve is: $y = a + bx^{-1} - 9x^{-2}$ where $a$ and $b$ are constants. We're also told that the maximum point on this curve is $(3, 4.4)$.

Since $(3, 4.4)$ is a maximum point, two conditions hold:

1. The function value at $x = 3$ should be $y = 4.4$.
2. The derivative $\frac{dy}{dx}$ at $x = 3$ should be zero (since it's a maximum point).

Step 1: Differentiate $y$ with respect to $x$

To find the derivative, $\frac{dy}{dx}$, we differentiate each term in the equation: $y = a + bx^{-1} - 9x^{-2}$

The derivative is: $\frac{dy}{dx} = 0 - b x^{-2} + 18x^{-3}$ or equivalently, $\frac{dy}{dx} = -\frac{b}{x^2} + \frac{18}{x^3}$

Step 2: Set $\frac{dy}{dx} = 0$ at $x = 3$

Since the maximum point occurs at $x = 3$, we set $\frac{dy}{dx} = 0$ when $x = 3$: $-\frac{b}{3^2} + \frac{18}{3^3} = 0$ Simplify: $-\frac{b}{9} + \frac{18}{27} = 0$ $-\frac{b}{9} + \frac{2}{3} = 0$

Multiply through by 9 to clear the fraction: $-b + 6 = 0$ $b = 6$

Step 3: Substitute $x = 3$, $y = 4.4$, and $b = 6$ into the original equation to solve for $a$

Now, substitute $x = 3$, $y = 4.4$, and $b = 6$ into the equation $y = a + bx^{-1} - 9x^{-2}$: $4.4 = a + 6 \cdot 3^{-1} - 9 \cdot 3^{-2}$

Calculate each term: $4.4 = a + 6 \cdot \frac{1}{3} - 9 \cdot \frac{1}{9}$ $4.4 = a + 2 - 1$ $4.4 = a + 1$

Solve for $a$: $a = 4.4 - 1$ $a = 3.4$

Final Answer

The value of $a$ is: $\boxed{3.4}$

Would you like further details on any of the steps?

Here are some related questions that might help deepen your understanding:

1. What is the significance of the derivative being zero at a maximum point?
2. How would you approach this problem if given only the minimum point instead of the maximum?
3. Why does substituting values of $x$ and $y$ into the original function help us solve for constants?
4. What role does the second derivative play in determining whether a point is a maximum?
5. Can this method be extended to solve for more complex functions?

Tip: When finding maxima or minima, remember to check both the first and second derivatives to confirm the nature of the critical points.