Let $f(x)=3 x-1$ and $g(x)=2-x$. Find the following.
(a) $(f+g)(x)$
(b) $(f-g)(x)$
(c) $(f \cdot g)(x)$
(d) $\frac{f}{g}$
(e) The domain of $\frac{f}{g}$
Answer (1)
Add the functions to find ( f + g ) ( x ) : ( f + g ) ( x ) = ( 3 x − 1 ) + ( 2 − x ) = 2 x + 1 .
Subtract the functions to find ( f − g ) ( x ) : ( f − g ) ( x ) = ( 3 x − 1 ) − ( 2 − x ) = 4 x − 3 .
Multiply the functions to find ( f \timges g ) ( x ) : ( f \timges g ) ( x ) = ( 3 x − 1 ) ( 2 − x ) = − 3 x 2 + 7 x − 2 .
Divide the functions to find g f ( x ) : g f ( x ) = 2 − x 3 x − 1 . The domain is all real numbers except where 2 − x = 0 , which is x = 2 . Thus, the domain is ( − ∞ , 2 ) ∪ ( 2 , ∞ ) . The final answer is ( − ∞ , 2 ) ∪ ( 2 , ∞ ) .
Explanation
Understanding the Problem We are given two functions, f ( x ) = 3 x − 1 and g ( x ) = 2 − x . Our goal is to find ( f + g ) ( x ) , ( f − g ) ( x ) , ( f \timges g ) ( x ) , g f ( x ) , and the domain of g f ( x ) . The problem provides the answers for ( f \timges g ) ( x ) and g f ( x ) , so we only need to verify them.
Finding (f+g)(x) To find ( f + g ) ( x ) , we add the two functions: ( f + g ) ( x ) = f ( x ) + g ( x ) = ( 3 x − 1 ) + ( 2 − x ) Simplifying the expression, we get: ( f + g ) ( x ) = 3 x − 1 + 2 − x = 2 x + 1 Thus, ( f + g ) ( x ) = 2 x + 1 .
Finding (f-g)(x) To find ( f − g ) ( x ) , we subtract the two functions: ( f − g ) ( x ) = f ( x ) − g ( x ) = ( 3 x − 1 ) − ( 2 − x ) Simplifying the expression, we get: ( f − g ) ( x ) = 3 x − 1 − 2 + x = 4 x − 3 Thus, ( f − g ) ( x ) = 4 x − 3 .
Finding (f \cdot g)(x) The problem states that ( f \timges g ) ( x ) = − 3 x 2 + 7 x − 2 . Let's verify this: ( f \timges g ) ( x ) = ( 3 x − 1 ) ( 2 − x ) = 3 x ( 2 − x ) − 1 ( 2 − x ) = 6 x − 3 x 2 − 2 + x = − 3 x 2 + 7 x − 2 Thus, ( f \timges g ) ( x ) = − 3 x 2 + 7 x − 2 is correct.
Finding f(x)/g(x) The problem states that g f ( x ) = 2 − x 3 x − 1 . This is simply the function f ( x ) divided by g ( x ) , so it is correct.
Finding the Domain of f(x)/g(x) To find the domain of g f ( x ) , we need to find the values of x for which the denominator is zero. We set g ( x ) = 2 − x = 0 and solve for x :
2 − x = 0 ⟹ x = 2 Thus, the domain of g f ( x ) is all real numbers except x = 2 . In interval notation, this is ( − ∞ , 2 ) ∪ ( 2 , ∞ ) .
Final Answer In summary: (a) ( f + g ) ( x ) = 2 x + 1 (b) ( f − g ) ( x ) = 4 x − 3 (c) ( f \timges g ) ( x ) = − 3 x 2 + 7 x − 2 (d) g f ( x ) = 2 − x 3 x − 1 (e) The domain of g f ( x ) is ( − ∞ , 2 ) ∪ ( 2 , ∞ ) .
Examples
Understanding function operations like addition, subtraction, multiplication, and division is crucial in many real-world applications. For example, in economics, if f ( x ) represents the cost of producing x items and g ( x ) represents the revenue from selling x items, then ( f − g ) ( x ) represents the profit. The domain of g f ( x ) would be important to determine the values of x for which the revenue is non-zero, ensuring that the profit calculation is valid. These concepts are also used in physics, engineering, and computer science to model and analyze various systems.
