For the differentiable function f: R - {0} ri | vedclass

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(D) Given the equation: $3 f(x) + 2 f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$  (Equation $1$)Replace $x$ with $\frac{1}{x}$: $3 f\left(\frac{1}{x}

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$13$

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(D) Given the equation: $3 f(x) + 2 f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$  (Equation $1$)Replace $x$ with $\frac{1}{x}$: $3 f\left(\frac{1}{x}\right) + 2 f(x) = x - 10$  (Equation $2$)Multiply Equation $1$ by $3$ and Equation $2$ by $2$:$9 f(x) + 6 f\left(\frac{1}{x}\right) = \frac{3}{x} - 30$$4 f(x) + 6 f\left(\frac{1}{x}\right) = 2x - 20$Subtracting the second from the first:$5 f(x) = \frac{3}{x} - 2x - 10$$f(x) = \frac{3}{5x} - \frac{2x}{5} - 2$Now,find $f(3)$:$f(3) = \frac{3}{5(3)} - \frac{2(3)}{5} - 2 = \frac{1}{5} - \frac{6}{5} - 2 = -1 - 2 = -3$Find $f^{\prime}(x)$:$f^{\prime}(x) = -\frac{3}{5x^2} - \frac{2}{5}$Calculate $f^{\prime}\left(\frac{1}{4}\right)$:$f^{\prime}\left(\frac{1}{4}\right) = -\frac{3}{5(1/16)} - \frac{2}{5} = -\frac{48}{5} - \frac{2}{5} = -\frac{50}{5} = -10$Finally,calculate $\left|f(3) + f^{\prime}\left(\frac{1}{4}\right)\right|$:$|-3 + (-10)| = |-13| = 13$

For the differentiable function $f: R - \{0\} \rightarrow R$,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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