We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
255
2
avatar+4116 

Help.

 Feb 2, 2018
 #1
avatar+7563 
+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+7563 
+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

7 Online Users