"For the differentiable function $f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$, let $3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10$, then $\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|$ is equal to"

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Accepted Answer

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Given the equation: $3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$
Replace $x$ with $\frac{1}{x}$ in the original equation:
$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$
Now, we have two equations:

$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$

$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$

By adding the two equations, we can find $f(x)$:

$5f(x) = \frac{3}{x} - 2x - 10$

Now, let's differentiate both sides with respect to $x$:

$5f'(x) = -\frac{3}{x^2} - 2$

Now, we can find the values for $f(3)$ and $f'\left(\frac{1}{4}\right)$:

$f(3) = \frac{1}{5}(1 - 6 - 10) = -3$

$f'\left(\frac{1}{4}\right) = \frac{1}{5}(-48 - 2) = -10$

Finally, calculate the expression we are interested in :

$\left|f(3) + f'\left(\frac{1}{4}\right)\right| = \left|-3 - 10\right| = 13$

"

For the differentiable function $f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$, let $3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10$, then $\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|$ is equal to

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For the differentiable function $f: \mathbb{R}-\{0\} \rightarrow \mathbb{R}$, let $3 f(x)+2 f\left(\frac{1}{x}\right)=\frac{1}{x}-10$, then $\left|f(3)+f^{\prime}\left(\frac{1}{4}\right)\right|$ is equal to

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