Fcn prob w/ restrictions: prove f(x) = ((3^x)+1) / (3^x) - (3^-x) >= 1 for x >= 0

[SamJohnson]

Fcn prob w/ restrictions: prove f(x) = ((3^x)+1) / (3^x) - (3^-x) >= 1 for x >= 0

I am genuinely stumped on this problem - it's for a pre-done exam for tomorrow.

Prove that, for all x values greater than zero, that the function f(x) = ((3^x)+1) / (3^x) - (3^-x) produces values greater than or equal to 1. Then, solve f(x) = 4.

Any help is greatly appreciated.

 

[Deleted member 4993]

I am genuinely stumped on this problem - it's for a pre-done exam for tomorrow.

Prove that, for all x values greater than zero, that the function f(x) = ((3^x)+1) / (3^x) - (3^-x) produces values greater than or equal to 1. Then, solve f(x) = 4.

Any help is greatly appreciated.

f(x) = [3^x + 1]/ 3^x - 3^(-x) = [3^x + 1] * [3^(-x)] - 3^(-x) = [3^(-x)] * [ 3^x + 1 - 1] ..... continue

 

[Ishuda]

I am genuinely stumped on this problem - it's for a pre-done exam for tomorrow.

Prove that, for all x values greater than zero, that the function f(x) = ((3^x)+1) / (3^x) - (3^-x) produces values greater than or equal to 1. Then, solve f(x) = 4.

Any help is greatly appreciated.

Assuming the expression is
f(x) = [ 3^x + 1 ] / [ 3^x - 3^(-x) ]
we have:
To make notation easier, let t = 3^x. We also note that, if x>0 then t>1. Restate the function
\(\displaystyle f(t)\, =\, \dfrac{t\, +\, 1}{t-\frac{1}{t}}\, =\, \dfrac{t\, [t\, +\, 1]}{t^2\, -\, 1}\)
Now continue by factoring the denominator and simplifying the function.