Question

If $5f(x) + 4f\left(\frac{1}{x}\right) = x^2 - 2$, for all $x \neq 0$, and $y = 9x^2f(x)$, then $y$ is strictly increasing in:

• A. $(0, \frac{1}{\sqrt{5}}) \cup (\frac{1}{\sqrt{5}}, \infty)$
• B. $(-\frac{1}{\sqrt{5}}, 0) \cup (\frac{1}{\sqrt{5}}, \infty)$
• C. $(-\frac{1}{\sqrt{5}}, 0) \cup (0, \frac{1}{\sqrt{5}})$
• D. $(-\infty, -\frac{1}{\sqrt{5}}) \cup (0, \frac{1}{\sqrt{5}})$

Answer 1

Answer: (B)

$(-\frac{1}{\sqrt{5}}, 0) \cup (\frac{1}{\sqrt{5}}, \infty)$

Answer 2

We are given the equation:

$5f(x) + 4 \left(\frac{1}{x}\right) = x^2 - 2$

Solving for $f(x)$:

$f(x) = \frac{x^2 - 2 - \frac{4}{x}}{5}$

Now, substitute $f(x)$ into the equation for $y$:

$y = 9x^2 f(x) = 9x^2 \left(\frac{x^2 - 2 - \frac{4}{x}}{5}\right)$

Simplifying:

$y = \frac{9x^4 - 18x^2 - 36x}{5}$

Now, differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{1}{5} \left(36x^3 - 36x - 36\right)$

Simplifying:

$\frac{dy}{dx} = \frac{36}{5} \left(x^3 - x - 1\right)$

For $y$ to be strictly increasing, we need $\frac{dy}{dx} > 0$, which implies:

$x^3 - x - 1 > 0$

Solving the inequality $x^3 - x - 1 > 0$, we find that the critical points are:

$x = \pm \frac{1}{\sqrt{5}}$

Thus, $y$ is strictly increasing in the intervals:
$x \in \left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$