Let $f(x)=x^2-7x$ and $g(x)=6+x$. Find the following:
(a) $(f+g)(x)$
(b) $(f - g)(x)$
(c) $(f \cdot g)(x)$
(d) $\left(\frac{f}{g}\right)(x)$
(e) The domain of $\frac{f}{g}$
(a) $(f+g)(x)=x^2-6x+6$ (Simplify your answer. Do not factor.)
(b) $( f - g )( x )=\square$ (Simplify your answer. Do not factor.)
Answer (2)
Subtract g ( x ) from f ( x ) to find ( f − g ) ( x ) = x 2 − 8 x − 6 .
Multiply f ( x ) and g ( x ) to find ( f ′ . ′ g ) ( x ) = x 3 − x 2 − 42 x .
Divide f ( x ) by g ( x ) to find ( g f ) ( x ) = 6 + x x 2 − 7 x .
Determine the domain of ( g f ) ( x ) by excluding x = − 6 , resulting in ( − ∞ , − 6 ) ∪ ( − 6 , ∞ ) .
x 2 − 8 x − 6
Explanation
Understanding the Problem We are given two functions, f ( x ) = x 2 − 7 x and g ( x ) = 6 + x . We need to find expressions for ( f − g ) ( x ) , ( f "." g ) ( x ) , and ( g f ) ( x ) , as well as the domain of ( g f ) ( x ) .
Calculating (f-g)(x) To find ( f − g ) ( x ) , we subtract g ( x ) from f ( x ) : ( f − g ) ( x ) = f ( x ) − g ( x ) = ( x 2 − 7 x ) − ( 6 + x ) Simplifying this expression, we get: ( f − g ) ( x ) = x 2 − 7 x − 6 − x = x 2 − 8 x − 6
Calculating (f . g)(x) To find ( f "." g ) ( x ) , we multiply f ( x ) by g ( x ) :
( f "." g ) ( x ) = f ( x ) ⋅ g ( x ) = ( x 2 − 7 x ) ( 6 + x ) Expanding this expression, we get: ( f "." g ) ( x ) = x 2 ( 6 + x ) − 7 x ( 6 + x ) = 6 x 2 + x 3 − 42 x − 7 x 2 = x 3 − x 2 − 42 x
Calculating (f/g)(x) To find ( g f ) ( x ) , we divide f ( x ) by g ( x ) :
( g f ) ( x ) = g ( x ) f ( x ) = 6 + x x 2 − 7 x This expression is already simplified.
Finding the Domain of (f/g)(x) To find the domain of ( g f ) ( x ) , we need to determine where the denominator g ( x ) = 6 + x is not equal to zero. We set 6 + x = 0 and solve for x :
6 + x = 0 ⟹ x = − 6 Therefore, the domain of ( g f ) ( x ) is all real numbers except x = − 6 . In interval notation, this is ( − ∞ , − 6 ) ∪ ( − 6 , ∞ ) .
Final Answer In summary: (a) ( f + g ) ( x ) = x 2 − 6 x + 6 (b) ( f − g ) ( x ) = x 2 − 8 x − 6 (c) ( f "." g ) ( x ) = x 3 − x 2 − 42 x (d) ( g f ) ( x ) = 6 + x x 2 − 7 x (e) The domain of g f is ( − ∞ , − 6 ) ∪ ( − 6 , ∞ ) .
Examples
Understanding function operations like addition, subtraction, multiplication, and division is crucial in many real-world applications. For instance, in business, if f ( x ) represents the revenue from selling x units of a product and g ( x ) represents the cost of producing x units, then ( f − g ) ( x ) gives the profit. The domain of g f is important because it tells us for what values of x the ratio of revenue to cost is meaningful, excluding any values that would lead to division by zero or other undefined situations. Analyzing these functions helps businesses make informed decisions about production levels and pricing strategies.
We found the following results: ( f + g ) ( x ) = x 2 − 6 x + 6 , ( f − g ) ( x ) = x 2 − 8 x − 6 , ( f ⋅ g ) ( x ) = x 3 − x 2 − 42 x , ( g f ) ( x ) = 6 + x x 2 − 7 x , and the domain of g f is ( − ∞ , − 6 ) ∪ ( − 6 , ∞ ) .
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