I have three questions:
1. Determine if the two functions f and g are inverses of each other algebraicially. If not why?
f(x)=2x+3/4x-3 and g(x)=3x+3/4x-2
A. no, (f * g)(x)= x+2/3
B. yes
C. no, (f * g)(x)=3x
2. Determine if the two functions f and g are inverses of each other algebraicially. If not why?
f(x)= -x^3 + 2 and g(x)= 3(on top of square root)(sqrt x-2/2)
A. no, (f * g)(x)= x-14/8
B. yes
C. no, (f * g)(x)= 3 - x/2
3. Determine if the two functions f and g are inverses of each other algebraicially. If not why?
f(x)=-2x+4/2-5x and g(x)=4-2x/5-2x
A. no, (f * g)(x)= -2x+6/3x-5
B. no, (f * g)(x)= -6x+6/3x-5
C. yes
+23251 I'll answer the first part; but I have two questions.
1) Are you using * to represent the composition of functions? Usually * asterisk) is used to indicate the multiplication of functions and º (a small circle) is used to represent the composition of functions.
2) I believe that the original problem has fractions as follows:
f(x) = (2x + 3) / (4x - 3)
g(x) = (3x + 3) / (4x - 2)
Are f(x) and g(x) inverse functions.
This can be solved in two ways:
1) Find the inverse of f(x) and see whether or not it equals g(x).
2) Find out if f( g(x) ) = g( f(x) ) = x
By the nature of the answers, I believe that the assignment was to use procedure #2.
(f º g)(x) = f( g(x) ) = [ 2 · g(x) + 3 ] / [ 4 · g(x) - 3 ]
= [ 2 · (3x + 3) / (4x - 2) + 3 ] / [ 4 · (3x + 3) / (4x - 2) - 3 ]
= [ (6x + 6) / (4x - 2) + 3 ] / [ (12x + 12) / (4x - 2) - 3 ]
Multiply both the numerator and the denominator by (4x - 2) and simplify:
= [ (6x + 6) + 3(4x - 2) ] / [ (12x + 12) - 3(4x - 2) ]
= [ 6x + 6 + 12x - 6 ] / [ 12x + 12 - 12x + 6 ]
= [ 18x ] / [ 18 ]
= x
I'll let you show that (g º f)(x) = x also.
This shows that the answer to problem 1 is 'yes'.
You can use this procedure with the other problems.
+23251 For problem #2, I'm going to write the cube root of x as x1/3 ...
So, I believe that g(x) is: g(x) = (x - 2)1/3 / 2
But, if f(x) = - x3 + 2,
then f( g(x) ) = - [ (x - 2)1/3 / 2 ] 3 + 2
Simplifying the cube:
= - (x - 2) / 8 + 2
= (-x + 2) / 8 + 2
= (-x + 2) / 8 + 16 / 8
= (-x + 2 + 16) / 8
= (-x + 18) / 8
However, if f(x) = x3 + 2,
then f( g(x) ) = [ (x - 2)1/3 / 2 ] 3 + 2
Simplifying the cube:
= (x - 2) / 8 + 2
= (x - 2) / 8 + 16 / 8
= (x - 2 + 16) / 8
= (x + 14) / 8
which would match answer (A).
The answer must be 'no', because the result isn't just 'x'.
Does this help?
Also, problem (3) seems to be like problem (1); can you follow the steps of problem (1) to finish problem (3)?
