For the pair of functions [tex]f(x)=\frac{5 x+1}{x-7}[/tex] and [tex]g(x)=\frac{6 x}{2 x-5}[/tex] find the following.
a. Find the functions [tex]f+g, f-g[/tex], fg, and [tex]\frac{f}{g}[/tex].
b. Determine the domain of the functions [tex]f+g, f-g[/tex], fg, and [tex]\frac{f}{g}[/tex].
Answer (2)
Find f + g : f ( x ) + g ( x ) = ( x − 7 ) ( 2 x − 5 ) 16 x 2 − 65 x − 5 .
Find f − g : f ( x ) − g ( x ) = ( x − 7 ) ( 2 x − 5 ) 4 x 2 + 19 x − 5 .
Find f g : f ( x ) g ( x ) = ( x − 7 ) ( 2 x − 5 ) 30 x 2 + 6 x .
Find g f : g ( x ) f ( x ) = 6 x 2 − 42 x 10 x 2 − 23 x − 5 .
The domain of f + g , f − g , and f g is ( − ∞ , 2 5 ) ∪ ( 2 5 , 7 ) ∪ ( 7 , ∞ ) .
The domain of g f is ( − ∞ , 0 ) ∪ ( 0 , 2 5 ) ∪ ( 2 5 , 7 ) ∪ ( 7 , ∞ ) .
f + g = ( x − 7 ) ( 2 x − 5 ) 16 x 2 − 65 x − 5 , f − g = ( x − 7 ) ( 2 x − 5 ) 4 x 2 + 19 x − 5 , f g = ( x − 7 ) ( 2 x − 5 ) 30 x 2 + 6 x , g f = 6 x 2 − 42 x 10 x 2 − 23 x − 5 and domains are as specified above.
Explanation
Understanding the problem We are given two functions, f ( x ) = x − 7 5 x + 1 and g ( x ) = 2 x − 5 6 x , and we need to find f + g , f − g , f g , g f and their domains.
Finding f+g First, we find f + g by adding the two functions: f ( x ) + g ( x ) = x − 7 5 x + 1 + 2 x − 5 6 x To add these fractions, we need a common denominator, which is ( x − 7 ) ( 2 x − 5 ) . So we have: ( x − 7 ) ( 2 x − 5 ) ( 5 x + 1 ) ( 2 x − 5 ) + ( x − 7 ) ( 2 x − 5 ) 6 x ( x − 7 ) = ( x − 7 ) ( 2 x − 5 ) ( 10 x 2 − 25 x + 2 x − 5 ) + ( 6 x 2 − 42 x ) = ( x − 7 ) ( 2 x − 5 ) 16 x 2 − 65 x − 5
Finding f-g Next, we find f − g by subtracting the two functions: f ( x ) − g ( x ) = x − 7 5 x + 1 − 2 x − 5 6 x Using the same common denominator, we have: ( x − 7 ) ( 2 x − 5 ) ( 5 x + 1 ) ( 2 x − 5 ) − ( x − 7 ) ( 2 x − 5 ) 6 x ( x − 7 ) = ( x − 7 ) ( 2 x − 5 ) ( 10 x 2 − 25 x + 2 x − 5 ) − ( 6 x 2 − 42 x ) = ( x − 7 ) ( 2 x − 5 ) 4 x 2 + 19 x − 5
Finding fg Now, we find f g by multiplying the two functions: f ( x ) g ( x ) = x − 7 5 x + 1 ⋅ 2 x − 5 6 x = ( x − 7 ) ( 2 x − 5 ) ( 5 x + 1 ) ( 6 x ) = ( x − 7 ) ( 2 x − 5 ) 30 x 2 + 6 x
Finding f/g Then, we find g f by dividing the two functions: g ( x ) f ( x ) = 2 x − 5 6 x x − 7 5 x + 1 = x − 7 5 x + 1 ⋅ 6 x 2 x − 5 = ( x − 7 ) ( 6 x ) ( 5 x + 1 ) ( 2 x − 5 ) = 6 x 2 − 42 x 10 x 2 − 23 x − 5
Finding the domain of f+g, f-g, and fg Now we determine the domain of f + g , f − g , and f g . The domain will be all real numbers except for the values of x that make the denominator of either f ( x ) or g ( x ) equal to zero. Thus, we solve x − 7 = 0 and 2 x − 5 = 0 to find these values. x − 7 = 0 ⟹ x = 7 and 2 x − 5 = 0 ⟹ x = 2 5 . Therefore, the domain of f + g , f − g , and f g is all real numbers except x = 7 and x = 2 5 . In interval notation, this is ( − ∞ , 2 5 ) ∪ ( 2 5 , 7 ) ∪ ( 7 , ∞ ) .
Finding the domain of f/g Finally, we determine the domain of g f . The domain will be all real numbers except for the values of x that make the denominator of either f ( x ) , g ( x ) , or the simplified expression of g f equal to zero. Thus, we solve x − 7 = 0 , 2 x − 5 = 0 , and 6 x = 0 to find these values. x − 7 = 0 ⟹ x = 7 , 2 x − 5 = 0 ⟹ x = 2 5 , and 6 x = 0 ⟹ x = 0 . Therefore, the domain of g f is all real numbers except x = 7 , x = 2 5 , and x = 0 . In interval notation, this is ( − ∞ , 0 ) ∪ ( 0 , 2 5 ) ∪ ( 2 5 , 7 ) ∪ ( 7 , ∞ ) .
Examples
Understanding function operations and their domains is crucial in many real-world applications. For instance, in physics, if f ( x ) represents the distance an object travels in time x and g ( x ) represents its velocity, then f ( x ) + g ( x ) could represent the combined effect of distance and velocity on another related quantity. Similarly, the domain restrictions ensure that we are only considering physically meaningful values of time, such as non-zero and finite intervals.
We found the functions f + g , f − g , f g , and g f with their respective denominators and represented them in simplified forms. The domains of f + g , f − g , f g exclude x = 7 and x = 2 5 , while the domain of g f also excludes x = 0 . Thus, their domains are provided in interval notation based on these exclusions.
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