# 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.