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avatar+4690 

Help.

 Feb 2, 2018
 #1
avatar+7348 
+2

3.

 

f ⋅ g   =   f(x) ⋅ g(x)

                                                Since  f(x)  =  -5x + 3  , we can replace  f(x)  with  -5x + 3

f ⋅ g   =   (-5x + 3) ⋅ g(x)

                                                Since  g(x)  =  6x - 2  , we can replace  g(x)  with  6x - 2

f ⋅ g   =   (-5x + 3) ⋅ (6x - 2)

 

Before we simplify, let's get the domain.

There are no denominators or anything, so the domain is all real numbers.

 

f ⋅ g   =   (-5x)(6x) + (-5x)(-2) + (3)(6x) + (3)(-2)

 

f ⋅ g   =  -30x2  +  10x  +  18x  +  -6           Combine like terms...

 

f ⋅ g   =  -30x2  +  28x  -  6

 Feb 2, 2018
 #2
avatar+7348 
+2

4.

 

\(\frac{f}{g}\,=\,\frac{f(x)}{g(x)} \\~\\ \frac{f}{g}\,=\,\frac{x^2-16}{x+4}\)

 

Before simplifying, let's get the domain.  x + 4  is in a denominator, so...

x + 4 ≠ 0

x  ≠  -4     The domain is all real numbers with the exception that  x ≠ -4

 

\(\frac{f}{g}\,=\,\frac{x^2-16}{x+4}\)

                                   Factor the numerator as a difference of squares.
\(\frac{f}{g}\,=\,\frac{(x-4)(x+4)}{x+4}\)

                                   Reduce the fraction by  (x + 4) .

\(\frac{f}{g}\,=\,\frac{(x-4)}{1} \\~\\ \frac{f}{g}\,=\,x-4\)

.
 Feb 2, 2018

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