Function Examples

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]

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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) = $1\over 2$($a^x – a^{-x}$).

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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.