1. Composite functions

Okay. Here it goes. I am wondering if I am doing the following correctly.

1) f(x) = 3x-2/-5x+7 ; g(x) = 7x-3/5x+6

Find f of g.

So I start by subbing them in so I get:

(3(7x-3/5x+6) -2) / (-5(7x-3/5x+6) + 7

I then multiply the 3 and -5 into the fractions:
((21x-9/5x+6)-2) / ((-35x+15/5x+6)+7)

Then I get a common denominator for the -2 and the 7:
((21x-9/5x+6) - (10x+12/5x+6)) / ((-35x+15/5x+6) + (35x+55/5x+6))

Then I subtract, add, and simplify to get:
11x-21/70

____________________

2) Find the inverse of f(x) = (3x-2/-5x+7)+5

I start by swapping the x's and y's:
x = (3y-2/-5y+7)+5

Then I multiply both sides to get:
-5xy+7x = (3y -2) +5

Then I swap the 7x and the 3y:
-5xy-3y = (-7x-2) + 5

Factor out the y:
y(-5x-3) = (-7x-2) + 5

Divide right side:
y = (-7x-2/-5x-3) + (5/-5x-3)

Add the -2 and 5 to get:
y = -7x+3/-5x-3

_____________

Are the above correct?

Thank you!

-Tom

2. Originally Posted by TomCat
Okay. Here it goes. I am wondering if I am doing the following correctly.

1) f(x) = 3x-2/-5x+7; g(x) = 7x-3/5x+6

Find f of g.

So I start by subbing them in so I get:

(3(7x-3/5x+6) -2) / (-5(7x-3/5x+6) + 7

I then multiply the 3 and -5 into the fractions:
((21x-9/5x+6)-2) / ((-35x+15/5x+6)+7)

Then I get a common denominator for the -2 and the 7:
((21x-9/5x+6) - (10x+12/5x+6)) / ((-35x+15/5x+6) + (35x+55/5x+6))

Then I subtract, add, and simplify to get:
11x-21/70

____________________

2) Find the inverse of f(x) = (3x-2/-5x+7)+5

I start by swapping the x's and y's:
x = (3y-2/-5y+7)+5

Then I multiply both sides to get:
-5xy+7x = (3y -2) +5
The first problem is correct. The second is wrong.

You multiplied both sides of the equation by (-5y+7). The RHS is thus:
[(3y-2)/(-5y+7) + 5] * (-5y+7) = (3y-2) + 5*(-5y+7) = -22y + 33

You should be able to take it from here.

A suggestion: Please be more careful in your use of parenthesis. The function:
f(x) = 3x-2/-5x+7 = 3x + 2/(5x) + 7 = 3x + 7 + 2/(5x)
is not defined for x = 0 and apparently not what you intended. You meant to write:
f(x) = (3x-2)/(-5x+7)

It sounds picky, but it could cost you points on a test.

-Dan

3. Ah... I see. I forgot to multiply it by the five. Okay. Thanks...

Noted! My Prof. really doesn't worry about the parentheses. At least, thus far he hasn't cared. I only use them on here to help split up the parts. I see what you mean though.

Thanks again!

-Tom

4. Originally Posted by TomCat
Ah... I see. I forgot to multiply it by the five. Okay. Thanks...

Noted! My Prof. really doesn't worry about the parentheses. At least, thus far he hasn't cared. I only use them on here to help split up the parts. I see what you mean though.

Thanks again!

-Tom
Your professor is lazy. You can tell him I said that.

It's really a good practice to be in. I recommend it even if your professor doesn't do it.

-Dan

5. Lol... Yeah. He's got something like seven classes four days a week with quizzes and tests every Thursday. I suppose that's his fault though. He has his own Intermediate Algebra book or something like that. Saldivar is his last name. He says how he wants to write a Pre-Calc book since he does a few things differently and in a totally different order, but he said it takes too much time. I just wish it wouldn't take so long to get the tests and quizzes back. That and he's already been sick a couple of times. We're not paying him to be sick, but apparently the Profs. get five sick days. I think we should back charge lol... That's life I suppose...

6. Am I doing this correctly...

Here are a few more problems... The last one I am stuck on how to simplify it correctly.

1) f(x) = ((2x-3)/(-7x+5)) +5 g(x) = (3x-7)/(6x+5)

Find f of g:

Put the g(x) in:
(2((3x-7)/(6x+5))-3) / (-7((3x-7)/(6x+5))+5) +5

((6x-14)/(6x+5)) -3 / (((-21x+49)/(6x+5)) +5) +5

Multiply the -3 and f by (6x+5)/(6x+5) and then add or subtract
((24x-29)/(6x+5) / (9x+74)/(6x+5)) +5

To get:
((24x-29)/(9x+74)) +5

I'm not sure if I dealt with the "+5" at the end of f(x) correctly...

________

2) Inverse of f(x) = ((2x-3)/(-7x+5)) +3

Multiply the 3 by (-7x+5)/(-7x+5)
y= ((2x-3)/(-7x+5)) + ((-21x+5)/(-7x+5))

That comes out to be:
y = (19x+12)/(-7x+5)

Switch the x's and y's:
x = (19y+12)/(-7y+5)

Multiply both sides by (-7y+5):
-7xy+5x = 19y+12

Switch the 5x with the 19y:
-7xy-19y = 12-5x

Factor out a y:
y(-7x-19) = 12-5x

Divide:
y = (12-5x)/(-7x-19)

__________

For this one, I'm not sure what I'm doing is called (sorry). My prof. has been doing them on the board and said to try a few different combinations at home. You should be able to pick up on what I'm doing, or trying to do anyway.

3) tan(cos^-1x + cot^-1y)

Missing triangle side is sqrt(1-x^2)

Missing side is sqrt((x^2) + 1)

tan(alpha + beta) = (tan alpha + tan beta)/(1- (tan alpha)(tan beta))

Sub in from the triangles:
(((sqrt(1-x^2))/(x)) +((1)/(x))) / (1- ((sqrt(1- x^2))/(x))((1)/(x))

How do I simplify that correctly?

Would it be:

(sqrt(1-x^2) +1) / (1 - (sqrt(1-x^2))

Thanks guys!

-Tom