Question

if the function f:R to R defined by f(x)=3^x+3^-x/2 then show that f(x+y)+f(x-y)=2f(x)f(y)
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Answer 1

Answer:

given, f(x)=3^x+3^-x/2

=>substitute x=x+y in f(x)

• f(x+y)=3^x+y+3^-(x+y)/2

substitute x=x-y in f(x)

• f(x-y)=3^(x-y)+3^-(x-y)/2

consider,f(x+y)+f(x-y)

=[3^(x+y)+3^-(x+y)/2]+[3^(x-y)+3^-(x-y)/2]

:(a^m x a^n=a^m+n)

=3^x x3^y+3^-x x3^-y+3^x x3^-y+3^-x x3^y/2

f(x+y)+f(x-y)=

={3^x x 3^y+3^-x x 3^y+3^x x 3^-y+3^-x x 3^y/2 }

• consider 2f(x).f(y)

=2f(x).f(y)= (3^x+3^-x/2)(3^y+3^-y/2)

2f(x).f(y)={3^x x 3^y+3^x x 3^-y+3^-x x 3^y+3^-x x 3^ -y/2}

• so, f(x+y)+f(x-y)=2f(x).f(y)
• Hence Proved

thankyou...

Answer 2

Answer:

I hope u will understand the following answer

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