First slide

Introduction to real valued functions

Question

$If 8 f (x) + 6 f (\frac{1}{x}) = x + 5 and y = x^{2} f (x), then \frac{d y}{d x} at x = - 1 is equal to$

Moderate

A

$\frac{1}{12}$
B

$\frac{1}{14}$
C

$\frac{19}{14}$
D

$- \frac{1}{14}$

Solution

$\begin{array}{l} given 8 f (x) + 6 f (\frac{1}{x}) = x + 5 . . . . . . . . . (1) \times 8 \\ Put x = \frac{1}{x} \Rightarrow 8 f (\frac{1}{x}) + 6 f (x) = \frac{1}{x} + 5 . . . . . . . (2) \times 6 \\ s o l v i n g 1 a n d 2 \\ From (1) & (2) f (x) = \frac{1}{28} (8 x - \frac{6}{x} + 10) \\ y = x^{2} f (x) \Rightarrow y = \frac{1}{28} (8 x^{3} - 6 x + 10 x^{2}) \\ \frac{d y}{d x} = \frac{1}{28} (24 x^{2} - 6 + 20 x) \\ {(\frac{d y}{d x})}_{x = - 1} = - \frac{1}{14} \end{array}$

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