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June 11th, 2017, 01:25 PM  #1 
Newbie Joined: May 2017 From: antartica Posts: 11 Thanks: 0  finding domain of square root function
State the domain of the following function using interval notation. sqrt( x^2 +10x 21) I got this answer (infinity, 5sqrt(46)] U [sqrt(46) 5, infinity) but it can't even be inputted in the answer part the answer part is as follows,(________,________) so I am not sure what to do. 
June 11th, 2017, 01:55 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,276 Thanks: 516 
The zeros of the function are $\displaystyle \sqrt{3}\ and\ \sqrt{7}$. The domain is that interval, since function is imaginary outside.

June 11th, 2017, 01:56 PM  #3  
Senior Member Joined: Sep 2015 From: CA Posts: 1,303 Thanks: 667  Quote:
$x^2  10x + 21 \leq 0$ $(x7)(x3) \leq 0$ $(x7)\leq 0 \wedge (x3)\geq 0 \bigvee (x7)\geq 0 \wedge (x3)\leq 0$ $x\leq 7 \wedge x\geq 3 \bigvee x \geq 7 \wedge x \leq 3$ the right hand condition is the empty set leaving $x \in [3, 7]$  
June 11th, 2017, 02:37 PM  #4 
Newbie Joined: May 2017 From: antartica Posts: 11 Thanks: 0 
That was correct thank you. I am still wondering though how you ignored the square root. the roots are 3 and 7 but what about thw square root do you just ignore it or...

June 11th, 2017, 02:53 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 2,611 Thanks: 1297  Quote:
$x^2+10x21 \ge 0$ ... as written by romsek. The values of x in the interval [3,7] satisfy this requirement ... all other values of x make the quadratic expression negative.  
June 11th, 2017, 02:59 PM  #6  
Newbie Joined: May 2017 From: antartica Posts: 11 Thanks: 0  Quote:
 

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