Let-f-be-a-function-defined-on-non-zero-real-numbers-such-that-27-f-x-x-x-2-f-1-x-2x-2-for-x-0-Find-f-x-f-3-

Question Number 116376 by bemath last updated on 03/Oct/20

Let f be a function defined on non zero

real numbers such that \frac{27 f (- x)}{x} - x^{2} f (\frac{1}{x}) = - {2 x}^{2}

for \forall x \neq 0 . Find \to {\begin{cases} f (x) \\ f (3) \end{cases} ?

Answered by bobhans last updated on 03/Oct/20

Letting x = - y, we get

- \frac{27 f (y)}{y} - y^{2} f (- \frac{1}{y}) = - {2 y}^{2} \dots (1)

Letting x = \frac{1}{y}, we get

27 y f (- \frac{1}{y}) - \frac{1}{y^{2}} f (y) = - \frac{2}{y^{2}} \dots (2)

Then 27 \times (1) + y \times (2) gives

- \frac{729 f (y)}{y} - \frac{f (y)}{y} = - {54 y}^{2} - \frac{2}{y}

solving for f (y), we have f (y) = \frac{1}{365} ({27 y}^{3} + 1)

or f (x) = \frac{{27 x}^{3} + 1}{365}, thus f (3) = \frac{3^{6} + 1}{365} = 2 .

Commented by bemath last updated on 03/Oct/20

santuyy

Answered by mathmax by abdo last updated on 03/Oct/20

change x by \frac{1}{x} we get 27 xf (- \frac{1}{x}) - \frac{1}{x^{2}} f (x) = - \frac{2}{x^{2}}

change x by - x weget - 27 xf (\frac{1}{x}) - \frac{1}{x^{2}} f (- x) = - \frac{2}{x^{2}} so

{\begin{cases} \frac{27}{x} f (- x) - x^{2} f (\frac{1}{x}) = - {2 x}^{2} \\ \frac{1}{x^{2}} f (- x) + 27 xf (\frac{1}{x}) = \frac{2}{x^{2}} \end{cases}

Δ_{s} = | \begin{matrix} \frac{27}{x} - x^{2} \\ \frac{1}{x^{2}} 27 x \end{matrix} | = 27^{2} + 1 \neq o \Rightarrow

f (- x) = \frac{| \begin{matrix} - {2 x}^{2} - x^{2} \\ \frac{2}{x^{2}} 27 x \end{matrix} |}{27^{2} + 1} = \frac{- {54 x}^{3} + 2}{27^{2} + 1} \Rightarrow

f (x) = \frac{{54 x}^{3} + 2}{27^{2} + 1} \Rightarrow f (3) = \frac{54 \times 3^{3} + 2}{27^{2} + 1}