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Let f(x) = (3x^2 - 10x - 25)/(x + 1) and g(x) = (3x^2 - 10x - 25)/(3x^2 + 11x + 10). Find the sum of all real numbers that are not in the domain of f(g(x)).

 

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If you could give me the answer to this question quickly, that would be very helpful. Thanks!

Guest Jan 29, 2018

Best Answer 

 #1
avatar+7179 
+3

\(f(\,g(x)\,)\,=\,\frac{3(\,g(x)\,)^2-\,10(\,g(x)\,)\,-\,25}{g(x)\,+\,1}\)

 

 

f( g(x) ) is undefined when  g(x)  is undefined, and  g(x)  is undefined when...

 

3x2 + 11x + 10  =  0

 

3x2 + 6x + 5x + 10  =  0

 

3x(x +  2) + 5(x + 2)  =  0

 

(x + 2)(3x + 5)  =  0

 

x  =  -2    and   x  =  -5/3

 

f( g(x) )  is also undefined when 

 

g(x) + 1  =  0

 

\(\frac{3x^2-10x-25}{3x^2+11x+10}+1\,=\,0 \\~\\ \frac{3x^2-10x-25}{3x^2+11x+10}+\frac{3x^2+11x+10}{3x^2+11x+10}\,=\,0 \\~\\ \frac{6x^2+x-15}{3x^2+11x+10}\,=\,0\)

 

6x2 + x - 15   =   0

 

6x2 - 9x + 10x - 15   =   0

 

3x(2x - 3) + 5(2x - 3)   =   0

 

(2x - 3)(3x + 5)   =   0

 

x  =  3/2   and   x = -5/3

 

f(g(x))  is undefined when  x = -2 ,  x = -5/3 ,  and  x = 3/2  .

 

-2  +  -5/3  +  3/2   =   -13/6

hectictar  Jan 29, 2018
 #1
avatar+7179 
+3
Best Answer

\(f(\,g(x)\,)\,=\,\frac{3(\,g(x)\,)^2-\,10(\,g(x)\,)\,-\,25}{g(x)\,+\,1}\)

 

 

f( g(x) ) is undefined when  g(x)  is undefined, and  g(x)  is undefined when...

 

3x2 + 11x + 10  =  0

 

3x2 + 6x + 5x + 10  =  0

 

3x(x +  2) + 5(x + 2)  =  0

 

(x + 2)(3x + 5)  =  0

 

x  =  -2    and   x  =  -5/3

 

f( g(x) )  is also undefined when 

 

g(x) + 1  =  0

 

\(\frac{3x^2-10x-25}{3x^2+11x+10}+1\,=\,0 \\~\\ \frac{3x^2-10x-25}{3x^2+11x+10}+\frac{3x^2+11x+10}{3x^2+11x+10}\,=\,0 \\~\\ \frac{6x^2+x-15}{3x^2+11x+10}\,=\,0\)

 

6x2 + x - 15   =   0

 

6x2 - 9x + 10x - 15   =   0

 

3x(2x - 3) + 5(2x - 3)   =   0

 

(2x - 3)(3x + 5)   =   0

 

x  =  3/2   and   x = -5/3

 

f(g(x))  is undefined when  x = -2 ,  x = -5/3 ,  and  x = 3/2  .

 

-2  +  -5/3  +  3/2   =   -13/6

hectictar  Jan 29, 2018
 #2
avatar+87677 
+1

Very nice, hectictar...!!!

 

cool cool cool

CPhill  Jan 29, 2018

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