Find f(a+b): $f(x)=\dfrac {2^x+2^{-x}}{2^x-2^{-x}}$

[Albert1]

$f(x)=\dfrac {2^x+2^{-x}}{2^x-2^{-x}},x\in R, x\neq 0$

$if$

$(1)f(a)=\dfrac{17}{15}$

$and$

$(2)f(b)=-\dfrac{65}{63}$

$find :f(a+b)$

 

[kaliprasad]

$f(x)=\dfrac {2^x+2^{-x}}{2^x-2^{-x}},x\in R, x\neq 0$

$if$

$(1)f(a)=\dfrac{17}{15}$

$and$

$(2)f(b)=-\dfrac{65}{63}$

$find :f(a+b)$

Spoiler

we have $f(a)=\frac {2^a+2^{-a}}{2^a-2^{-a}} = \frac{2^{2a}+1}{2^{2a}-1} = 1 + \frac{2}{2^{2a}-1} = 1 + \frac{2}{15}$
hence $2^{2a}-1 = 15$ or $a = 2$
we have $f(b)=\frac {2^b+2^{-b}}{2^b-2^{-b}} = \frac{1 + 2^{-2b}}{1 - 2^{-2b}} = \frac{-65}{63}$
or
$\frac{1 + 2^{-2b}}{2^{-2b}-1} = \frac{-65}{63}$ giving -2b = 6(same method as for a) or b = -3

so a + b = -1

so $f(a+b) = \frac{2^{-1} + 2^1}{2^{-1} - 2^1} = \frac{-5}{3}$

 

[Monoxdifly]

$f(x)=\dfrac {2^x+2^{-x}}{2^x-2^{-x}},x\in R, x\neq 0$

$if$

$(1)f(a)=\dfrac{17}{15}$

$and$

$(2)f(b)=-\dfrac{65}{63}$

$find :f(a+b)$

Spoiler

$f(a)=\dfrac{17}{15}$
$\dfrac {2^a+2^{-a}}{2^a-2^{-a}}=\dfrac{17}{15}$
$\dfrac {2^{2a}+1}{2^{2a}-1}=\dfrac{17}{15}$
$17(2^{2a}-1)=15(2^{2a}+1)$
$17\cdot2^{2a}-17=15\cdot2^2a+15$
$2\cdot2^{2a}=32$
$2^{2a}=16$
$2a=4$
$a=2$

$f(b)=-\dfrac{65}{63}$
$\dfrac {2^b+2^{-b}}{2^b-2^{-b}}=-\dfrac{65}{63}$
$\dfrac {2^{2b}+1}{2^{2b}-1}=-\dfrac{65}{63}$
$65(2^{2b}-1)=-63(2^{2b}+1)$
$65\cdot2^{2b}-65=-63\cdot2^2b-63$
$128\cdot2^{2b}=2$
$2^{2b}=\dfrac{2}{128}$
$2^{2b}=\dfrac{1}{64}$
$2^{2b}=2^{-6}$
$2b=-6$
$b=-3$

$f(a+b)=f(2-3)$
$f(a+b)=f(-1)$
$f(a+b)=\dfrac {2^{-1}+2^{1}}{2^{-1}-2^{1}}$
$f(a+b)=\dfrac {\frac{5}{2}}{-\frac{3}{2}}$
$f(a+b)=-\dfrac {5}{3}$