Math, asked by shradi2005, 7 days ago

find the domain of the following real valued function. ​

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Answered by harshsharma88494
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].

And here is a notice (and request) for you, My friend...

If you have more interesting mathematical problems, then please mail me at:-

[email protected]

I'm a 10th graded student and I love learning mathematics. So I keep learning new topics. So please email me your problems.

This will be helpful for both of us.

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Answered by Anonymous
8

 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}  &gt; 0 \\ 3 + x &gt; 0 \\ x &gt;  - 3 \\ domain = ( - 3, \:  \infty )....................(2)  \\  2. \sqrt{3 - x}  &gt; 0 \\ 3 - x &gt; 0 \\  x - 3 &lt; 0 \\ x &lt; 3 \\ domain = (  - \infty , \: 3).....................(3) \\ by \: taking \: intersection  \\ \: of \: (1) \: (2) \: and \: (3) \\ x = ( - 3 ,\: 3) \:

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