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...

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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)

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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)

cosx= x/1 = adj/hyp

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

cotx= x/1 = adj/opp

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