Key Concepts and Formulas

• Functional Equations: Substituting $x$ with $\frac{1}{x}$ in a functional equation involving $f(x)$ and $f(\frac{1}{x})$ can create a system of equations solvable for $f(x)$.
• Strictly Increasing Function: A differentiable function $y(x)$ is strictly increasing on an interval if $y'(x) > 0$ on that interval.
• Quotient Rule: The derivative of $\frac{u(x)}{v(x)}$ is $\frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$.

Step-by-Step Solution

Step 1: Solve for $f(x)$

We are given the functional equation: $5f(x) + 4f\left(\frac{1}{x}\right) = x^2 - 2 \quad (*)$ Replace $x$ with $\frac{1}{x}$ in the equation: $5f\left(\frac{1}{x}\right) + 4f\left(\frac{1}{\frac{1}{x}}\right) = \left(\frac{1}{x}\right)^2 - 2$ $5f\left(\frac{1}{x}\right) + 4f(x) = \frac{1}{x^2} - 2 \quad (**)$ Now we have a system of two equations with two unknowns, $f(x)$ and $f(\frac{1}{x})$. We can solve for $f(x)$. Multiply equation $(*)$ by 5 and equation $(**)$ by 4: $25f(x) + 20f\left(\frac{1}{x}\right) = 5x^2 - 10$ $16f(x) + 20f\left(\frac{1}{x}\right) = \frac{4}{x^2} - 8$ Subtract the second equation from the first: $(25-16)f(x) = (5x^2 - 10) - \left(\frac{4}{x^2} - 8\right)$ $9f(x) = 5x^2 - 10 - \frac{4}{x^2} + 8$ $9f(x) = 5x^2 - 2 - \frac{4}{x^2}$ $f(x) = \frac{1}{9}\left(5x^2 - 2 - \frac{4}{x^2}\right)$

Step 2: Find $y(x)$

We are given $y = 9x^2 f(x)$. Substitute the expression for $f(x)$ we found in Step 1: $y = 9x^2 \cdot \frac{1}{9}\left(5x^2 - 2 - \frac{4}{x^2}\right)$ $y = x^2 \left(5x^2 - 2 - \frac{4}{x^2}\right)$ $y = 5x^4 - 2x^2 - 4$

Step 3: Find $y'(x)$

To find the intervals where $y(x)$ is strictly increasing, we need to find $y'(x)$ and determine where it is positive. $y'(x) = \frac{d}{dx}(5x^4 - 2x^2 - 4)$ $y'(x) = 20x^3 - 4x$ $y'(x) = 4x(5x^2 - 1)$

Step 4: Determine where $y'(x) > 0$

We want to find where $4x(5x^2 - 1) > 0$. This is equivalent to $x(5x^2 - 1) > 0$. We can rewrite this as $x(\sqrt{5}x - 1)(\sqrt{5}x + 1) > 0$. The roots of $y'(x) = 0$ are $x = 0$, $x = \frac{1}{\sqrt{5}}$, and $x = -\frac{1}{\sqrt{5}}$.

Consider the intervals:

• $x < -\frac{1}{\sqrt{5}}$: $x < 0$, $\sqrt{5}x - 1 < 0$, $\sqrt{5}x + 1 < 0$. So $x(\sqrt{5}x - 1)(\sqrt{5}x + 1) < 0$.
• $-\frac{1}{\sqrt{5}} < x < 0$: $x < 0$, $\sqrt{5}x - 1 < 0$, $\sqrt{5}x + 1 > 0$. So $x(\sqrt{5}x - 1)(\sqrt{5}x + 1) > 0$.
• $0 < x < \frac{1}{\sqrt{5}}$: $x > 0$, $\sqrt{5}x - 1 < 0$, $\sqrt{5}x + 1 > 0$. So $x(\sqrt{5}x - 1)(\sqrt{5}x + 1) < 0$.
• $x > \frac{1}{\sqrt{5}}$: $x > 0$, $\sqrt{5}x - 1 > 0$, $\sqrt{5}x + 1 > 0$. So $x(\sqrt{5}x - 1)(\sqrt{5}x + 1) > 0$.

Thus, $y'(x) > 0$ when $-\frac{1}{\sqrt{5}} < x < 0$ or $x > \frac{1}{\sqrt{5}}$. Therefore, $y(x)$ is strictly increasing on the intervals $\left(-\frac{1}{\sqrt{5}}, 0\right)$ and $\left(\frac{1}{\sqrt{5}}, \infty\right)$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when solving inequalities. A sign error can easily lead to the wrong interval.
• Don't Forget the Domain: Always consider the domain of the original function. In this case, $x \neq 0$.
• Factoring: Factoring the derivative can simplify the process of finding the intervals where it is positive or negative.

Summary

We solved the functional equation for $f(x)$, substituted it into the expression for $y(x)$, found the derivative $y'(x)$, and determined the intervals where $y'(x) > 0$. This gave us the intervals where $y(x)$ is strictly increasing. The final answer is $\left(0, \frac{1}{\sqrt{5}}\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$.

Final Answer

The final answer is $\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$, which corresponds to option (B).