Solution :
Given f(x) = \(log_a(x + \sqrt{(x^2+1)})\)
f'(x) = \(log_ae\over {\sqrt{1+x^2}}\) > 0
which is strictly increasing functions.
Thus, f(x) is injective, given that f(x) is onto. Hence the given function f(x) is invertible.
Interchanging x & y
\(\implies\) \(log_a(y + \sqrt{(y^2+1)})\) = x
\(\implies\) \(y + \sqrt{(y^2+1)}\) = \(a^x\) ……..(1)
and \(\sqrt{(y^2+1)}\) – y = \(a^{-x}\) ………..(2)
From (1) and (2), we get y = \(1\over 2\)(\(a^x – a^{-x}\)) or \(f{-1}\)(x) = \(1\over 2\)(\(a^x – a^{-x}\)).
Similar Questions
If y = 2[x] + 3 & y = 3[x – 2] + 5, then find [x + y] where [.] denotes greatest integer function.
Find the domain and range of function f(x) = \(x-2\over 3-x\).
Find the period of the function f(x) = \(e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}\)
Find the domain of the function f(x) = \(1\over x + 2\).
Find the range of the function \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\)
