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Question: If $8f(x)+6f(\frac{1}{x})=x+5$ and $y = x^2 f(x)$, then $\frac{dy}{dx}$ at $x = -1$ is...

If 8f(x)+6f(1x)=x+58f(x)+6f(\frac{1}{x})=x+5 and y=x2f(x)y = x^2 f(x), then dydx\frac{dy}{dx} at x=−1x = -1 is

A

-14

B

14

C

-\frac{1}{14}

D

\frac{1}{14}

Answer

-\frac{1}{14}

Explanation

Solution

The given functional equation is solved to find f(x)f(x) by treating it as a system of linear equations. yy is then expressed as a polynomial in xx. Differentiation of yy and evaluation at x=−1x=-1 yields the result.