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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

Guest Sep 16, 2016
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4+0 Answers

 #1avatar+23252 0

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.

geno3141 Sep 16, 2016 #2avatar0

Yeah that's what i'm trying to use

Guest Sep 16, 2016 #3avatar0

I don't get the other problems..

Guest Sep 16, 2016 #4avatar+23252 0

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)?

geno3141 Sep 16, 2016 Post New Answer

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