Function Examples

FUNCTION EXAMPLES

Example 1 : Find the range of the 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.

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