Domain of function?

**offahengaway and chips** · September 22nd 2008, 06:39 AM

Is the domain of

f(x)= 1/sqrt(6-x)(5+x)

(-5,6) or (5,6) ?

**Shyam** · September 22nd 2008, 07:20 AM

Originally Posted by offahengaway and chips

Is the domain of

f(x)= 1/sqrt(6-x)(5+x)

(-5,6) or (5,6) ?

$f\left( x \right) = \frac{1} {{\sqrt {\left( {6 - x} \right)\left( {5 + x} \right)} }}$

For domain, the term inside the square-root should be non-negative.

so, $(6-x)(5+x)\ge 0$

therefore, $6-x \ge 0 \;\;and \;\; 5+x \ge 0$

$\Rightarrow 6 \ge x \;\; and \;\; x \ge -5$

$\Rightarrow -5 \le x \le 6$

$\Rightarrow x \in (-5, 6)$

Also, $6-x \le 0 \;\;and \;\; 5+x \le 0$

$\Rightarrow x \le -5 \;\; and \;\; x \ge 6$

$\Rightarrow No \;\;Solution$

**Krizalid** · September 22nd 2008, 05:55 PM

That has to be wrong, 'cause we require that $(6-x)(5+x)>0.$ After some simple calculations we have,

$\begin{aligned} 30+x-x^{2}&>0 \\ x^{2}-x-30&<0 \\ (2x-1)^{2}-121&<0 \\ -11<2x-1&<11, \end{aligned}$

and we happily get $-10<2x<12\implies -5<x<6,$ which is the required domain.

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