Here you will learn some function examples for better understanding of function concepts.
Example 1 : Find the range of the given function \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\)
Solution : Now 1 \(\le\) \(16sin^2x\) + 1) \(\le\) 17
\(\therefore\) 0 \(\le\) \(log_2(16sin^2x+1)\) \(\le\) \(log_217\)
\(\therefore\) 2 – \(log_217\) \(\le\) 2 – \(log_2(16sin^2x+1)\) \(\le\) 2
Now consider 0 < 2 – \(log_2(16sin^2x+1)\) \(\le\) 2
\(\therefore\) -\(\infty\) < \(log_{\sqrt{2}}(2-log_2(16sin^2x+1))\) \(\le\) \(log_{\sqrt{2}}2\) = 2
\(\therefore\) the range is (-\(\infty\), 2]
Example 2 : Find the inverse of the function f(x) = \(log_a(x + \sqrt{(x^2+1)})\); a > 1 and assuming it to be an onto function.
Solution : Given f(x) = \(log_a(x + \sqrt{(x^2+1)})\)
\(\therefore\) 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) = \(1\over 2\)(\(a^x – a^{-x}\)).
Example 3 : Find the period of the function f(x) = \(e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}\)
Solution : f(x) = \(e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}\)
Period of x – [x] = 1
Period of \(|cos\pi x|\) = 1
Period of \(|cos2\pi x|\) = \(1\over 2\)
……………………………….
Period of \(|cosn\pi x|\) = \(1\over n\)
So period of f(x) will be L.C.M of all period = 1.
Practice these given function examples to test your knowledge on concepts of function.