The problem in the image asks to find the minimum value of the function:

$y = 2(3^x + 3^{-x}) - (9^x + 9^{-x}) + 5$

Step 1: Simplify the expression

We can express $9^x$ and $9^{-x}$ in terms of $3^x$ and $3^{-x}$ because $9 = 3^2$. Thus:

$9^x = (3^2)^x = (3^x)^2$ $9^{-x} = (3^{-2})^x = (3^{-x})^2$

Now, substitute into the original equation:

$y = 2(3^x + 3^{-x}) - ((3^x)^2 + (3^{-x})^2) + 5$

Step 2: Let $a = 3^x + 3^{-x}$

This is a common substitution to simplify the expression. So we now have:

$y = 2a - (a^2 - 2) + 5$

The $-2$ comes from the identity: $(3^x + 3^{-x})^2 = (3^x)^2 + (3^{-x})^2 + 2$.

Step 3: Simplify further

$y = 2a - a^2 + 2 + 5$ $y = -a^2 + 2a + 7$

Step 4: Find the minimum value of $y$

The function $y = -a^2 + 2a + 7$ is a quadratic function in $a$. Since the coefficient of $a^2$ is negative, the parabola opens downward, and the maximum value can be found at the vertex.

The vertex of a quadratic function $y = -a^2 + 2a + 7$ occurs at:

$a = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1$

Step 5: Calculate the minimum value of $y$

Substitute $a = 1$ into the equation for $y$:

$y = -(1)^2 + 2(1) + 7$ $y = -1 + 2 + 7 = 8$

Thus, the minimum value of the function is 8.

Final Answer:

The correct answer is $\boxed{8}$.

Would you like more details, or do you have any questions?

Here are 5 related questions to explore further:

1. How do you derive the identity $(3^x + 3^{-x})^2 = (3^x)^2 + (3^{-x})^2 + 2$?
2. How can you apply vertex formulas to solve other quadratic equations?
3. How can substitution simplify solving exponential expressions?
4. What are the properties of quadratic functions that open downwards?
5. How do transformations of the function $3^x$ affect its graph?

Tip: When solving problems involving exponential terms, always check if a substitution can simplify expressions.