Question

find the domain of the following real valued function. ​

Answer 1

Answer:

It's just a game of thinking...

Step-by-step explanation:

$f(x) = \frac{ \sqrt{3 + x} + \sqrt{3 - x} }{x}$

Take a look at the expression:-

$\sqrt{3 + x}$

• Which Implies that x can't be lesser than -3.

Now, look at this expression :-

$\sqrt{3 - x}$

Which Implies that x can't be greater than 3.

So, the domain of f (x) should be [-3, 3].

But wait !

Checked the x in the denominator?

What does this x mean?

(This means that x can't be equal to 0, as f (x) will approach to infinite).

So, finally...

The domain of f (x) is [-3, 0) U (0, 3].

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Answer 2

$f(x) = \frac{ \sqrt{3 + x } \: + \sqrt{3 - x} }{x}$

$\:$

$To \: define \: the \: equation \\ </p><p> denominator \: cannot \: be \:zero$

$∴ \: x \: cannot \: be \: zero....(1)</p><p>$

${ \sqrt{3 + x } + \sqrt{3 - x} }{} \\ \\ for \: root \: to \: be \: defined \: \\ 1. \sqrt{3 + x} \geqslant 0 \: \\ but \: it \: cant \: be \: 0 \\ \: becoz \: of \: denomintaor \\ so \: \sqrt{3 + x} > 0 \\ 3 + x > 0 \\ x > - 3 \\ domain = ( - 3, \: \infty )....................(2) \\ 2. \sqrt{3 - x} > 0 \\ 3 - x > 0 \\ x - 3 < 0 \\ x < 3 \\ domain = ( - \infty , \: 3).....................(3) \\ by \: taking \: intersection \\ \: of \: (1) \: (2) \: and \: (3) \\ x = ( - 3 ,\: 3) \:$