BeMath-Given-2f-x-3f-1-x-2x-3-find-f-3-

Question Number 108193 by bemath last updated on 15/Aug/20

\frac{♡ B e M a t h ♡}{⧫}

G i v e n 2 f (x) + 3 f (\frac{1}{x}) = 2 x + 3

f i n d f (3) ?

Commented by bemath last updated on 15/Aug/20

t h a n k y o u b o t h

Answered by john santu last updated on 15/Aug/20

\frac{✓ JS ✓}{♠}

p u t x = 3 \Rightarrow 2 f (3) + 3 f (\frac{1}{3}) = 9

p u t x = \frac{1}{3} \Rightarrow 2 f (\frac{1}{3}) + 3 f (3) = \frac{2}{3} + 3

{\begin{cases} 4 f (3) + 6 f (\frac{1}{3}) = 18 \\ 9 f (3) + 6 f (\frac{1}{3}) = 11 \end{cases}

(1) - (2) \Rightarrow - 5 f (3) = 7

\Rightarrow f (3) = - \frac{7}{5}

Answered by mr W last updated on 15/Aug/20

2 f (x) + 3 f (\frac{1}{x}) = 2 x + 3 \dots (i)

r e p l a c e x w i t h \frac{1}{x}

2 f (\frac{1}{x}) + 3 f (x) = \frac{2}{x} + 3 \dots (i i)

3 (i i) - 2 (i) :

5 f (x) = 3 + \frac{6}{x} - 4 x

\Rightarrow f (x) = \frac{3}{5} + \frac{6}{5 x} - \frac{4 x}{5}

\Rightarrow f (3) = - \frac{7}{5}

Answered by Dwaipayan Shikari last updated on 15/Aug/20

2 f (3) + 3 f (\frac{1}{3}) = 9 \Rightarrow 4 f (3) + 6 f (\frac{1}{3}) = 18

3 f (3) + 2 f (\frac{1}{3}) = \frac{11}{3} \Rightarrow 9 f (3) + 6 f (\frac{1}{3}) = 11

- 5 f (3) = 7

f (3) = - \frac{7}{5}