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Mathematics
Evaluate ( f g ) ( x ) (f@g)(x) and write the domain in interval notation. Write the answer in the intervals as an integer or simplified fraction.
f ( x ) = x x 1 g ( x ) = 13 x 2 36 f(x)=(x)/(x-1)quad g(x)=(13)/(x^(2)-36)
Part: 0 / 2 0//2
Part 1 of 2
( f g ) ( x ) = (f@g)(x)=
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To evaluate ( f g ) ( x ) (f@g)(x), we need to first find g ( x ) g(x) and then evaluate f f at g ( x ) g(x).
Given that f ( x ) = x x 1 f(x)=(x)/(x-1) and g ( x ) = 13 x 2 36 g(x)=(13)/(x^(2)-36), we have
( f g ) ( x ) = f ( g ( x ) ) = f ( 13 x 2 36 ) = 13 x 2 36 13 x 2 36 1 . (f@g)(x)=f(g(x))=f((13)/(x^(2)-36))=((13)/(x^(2)-36))/((13)/(x^(2)-36)-1).
Simplifying the denominator, we get
( f g ) ( x ) = 13 x 2 36 13 x 2 36 1 x 2 36 x 2 36 = 13 ( x 2 36 ) 13 = 13 x 2 49 = 13 ( x 7 ) ( x + 7 ) {:[(f@g)(x)=((13)/(x^(2)-36))/((13)/(x^(2)-36)-1)*(x^(2)-36)/(x^(2)-36)],[],[=(13)/((x^(2)-36)-13)],[],[=(13)/(x^(2)-49)],[],[=(13)/((x-7)(x+7))]:}
Therefore, ( f g ) ( x ) = 13 ( x 7 ) ( x + 7 ) (f@g)(x)=(13)/((x-7)(x+7)).
Now, let’s find the domain of ( f g ) ( x ) (f@g)(x). The function is undefined when the denominator ( x 7 ) ( x + 7 ) (x-7)(x+7) equals zero. This happens when x = 7 x=7 or x = 7 x=-7. So, the domain of ( f g ) ( x ) (f@g)(x) is all real numbers except for x = 7 x=7 and x = 7 x=-7.
In interval notation, the domain of ( f g ) ( x ) (f@g)(x) is ( , 7 ) ( 7 , 7 ) ( 7 , ) (-oo,-7)uu(-7,7)uu(7,oo).
Answer: ( f g ) ( x ) = 13 ( x 7 ) ( x + 7 ) (f@g)(x)=(13)/((x-7)(x+7)) with domain ( , 7 ) ( 7 , 7 ) ( 7 , ) (-oo,-7)uu(-7,7)uu(7,oo).
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