The function given in the image is:

$y = -3 \cdot (x - b)^2 - (x - a)^2 + 9$

To find the values of $x$ that maximize or minimize this function, we'll need to take the derivative of the function $y(x)$ with respect to $x$ and set it equal to zero to find the critical points.

Step-by-Step Solution:

1. Differentiate $y(x)$ with respect to $x$:

$y(x) = -3 \cdot (x - b)^2 - (x - a)^2 + 9$

Using the chain rule, we differentiate each term:

$\frac{d}{dx} \left( -3 \cdot (x - b)^2 \right) = -6 \cdot (x - b)$ $\frac{d}{dx} \left( - (x - a)^2 \right) = -2 \cdot (x - a)$ The constant 9 vanishes, as its derivative is zero.

Thus, the derivative is:

$\frac{dy}{dx} = -6 \cdot (x - b) - 2 \cdot (x - a)$

2. Set the derivative equal to zero to find the critical points:

$-6 \cdot (x - b) - 2 \cdot (x - a) = 0$

This simplifies to:

$-6x + 6b - 2x + 2a = 0$ $-8x + 6b + 2a = 0$

3. Solve for $x$:

$8x = 6b + 2a$ $x = \frac{6b + 2a}{8} = \frac{3b + a}{4}$

Conclusion:

The value of $x$ that either maximizes or minimizes the function is:

$x = \frac{3b + a}{4}$

This is the critical point. To determine whether it gives a maximum or minimum, you would check the second derivative, but given the structure of the function (with negative quadratic terms), this should correspond to a maximum.

Would you like further details on the second derivative test or other aspects of this solution?

Here are 5 follow-up questions based on this topic:

1. How would you apply the second derivative test to confirm if this critical point is a maximum or minimum?
2. Can you generalize this method for other quadratic functions?
3. How does the presence of multiple squared terms affect the nature of critical points in this function?
4. What happens if the coefficients in the equation change from -3 and -1 to positive values?
5. How can you interpret the physical meaning of a maximum or minimum in real-world applications?

Tip: Always verify whether a critical point is a maximum or minimum using the second derivative test for conclusive results.