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Mathematics 15 Online
OpenStudy (israkjahim):

Can someone help me please Part 1. Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses. f(x)=x+a/b g(x)=cx-d Part 2. Show your work to prove that the inverse of f(x) is g(x). Part 3. Show your work to evaluate g(f(x)). Part 4. Graph your two functions on a coordinate plane. Include a table of values for each function. Include 5 values for each function. Graph the line y = x on the same graph.

OpenStudy (israkjahim):

can someone help me please

OpenStudy (phi):

You should pick simple numbers for a and b

OpenStudy (israkjahim):

a is 6 and b is 9

OpenStudy (phi):

by the way is it \[ f(x) = x + \frac{a}{b} \] or \[ f(x) = \frac{x+a}{b} \]

OpenStudy (phi):

also, life is a bit easier if you use *small numbers* like 1 and 2!

OpenStudy (israkjahim):

we could use the second one

OpenStudy (israkjahim):

sure

OpenStudy (phi):

It makes a difference to the answer. can you use the equation editor to post the exact equations for f(x) and g(x) ? otherwise, we do a lot of work and get the wrong answer.

OpenStudy (israkjahim):

f(x)=x+1/2

OpenStudy (phi):

and is that x plus one-half or (x+1) all divided by 2 ?

OpenStudy (israkjahim):

\[f(x)=\frac{ x+a }{ b}\]

OpenStudy (israkjahim):

\[g(x)=cx-d\]

OpenStudy (phi):

ok, so you should write it as f(x) = (x+1)/2 to find the inverse, rename f(x) as y y= (x+1)/2 (and remember this , because they want you to plot this line) next, "swap the x and y" can you do that ?

OpenStudy (phi):

swap means erase the "y" and put x in its place then erase the original "x" and put in y what do you get ?

OpenStudy (israkjahim):

y=1/2+1 i guess

OpenStudy (phi):

start with y= (x+1)/2 change the first "y" into x what do you get ?

OpenStudy (john_doe02125):

Is this the segment honors project?

OpenStudy (israkjahim):

yes

OpenStudy (phi):

then in the (x+1) change that "x" into a y

OpenStudy (israkjahim):

i confused

OpenStudy (phi):

the procedure to find the inverse has a few steps the first step is to "rename" x to y and the y to x there is no thinking involved. If you see y, write x, if you see x write y start with y = (x+1)/2

OpenStudy (phi):

and leave all the other "stuff" as it is

OpenStudy (israkjahim):

x=(y+1)/2

OpenStudy (phi):

yes

OpenStudy (phi):

the next step is "solve for y" I would start by multiplying both sides by 2 that means write *2 on both sides (that means times 2)

OpenStudy (israkjahim):

x(y+1)/4

OpenStudy (phi):

it is clearer if you write it like this \[ x = \frac{y+1}{2} \\ 2\cdot x = 2\cdot \frac{y+1}{2} \]

OpenStudy (phi):

in algebra we don't usually show the multiply sign when we do number times letter we would write 2*x as just 2x

OpenStudy (phi):

on the right side you have a 2 "up top" and a 2 "down below" that means you have \[ \frac{2}{2} \] do you know what that simplifies to ?

OpenStudy (israkjahim):

1

OpenStudy (phi):

yes, so the right side is the same as 1*(y+1) which is just (y+1) so far you have \[ 2x= (y+1)\] the parens are not doing anything so it is the same as \[ 2x = y+1 \] or \[ y+1= 2x \]

OpenStudy (phi):

next, write -1 on both sides what do you get ?

OpenStudy (israkjahim):

2(1)=1+1

OpenStudy (phi):

start with y+1 = 2x write -1 on each side don't do anything else. what does it look like?

OpenStudy (israkjahim):

1+1=2x

OpenStudy (israkjahim):

1+=2(1)

OpenStudy (phi):

no, we leave the y alone you have y+1 = 2x now write -1 on both sides, like this y+1 -1 = 2x -1

OpenStudy (phi):

on the left side you have 1 take away 1

OpenStudy (phi):

1 -1 is 0 y+0 is y (adding 0 does not do anything) you're answer is y= 2x-1 the last step is rename y to be g(x)

OpenStudy (phi):

g(x)= 2x - 1 now you know what 'c" and "d" are, by matching your answer with g(x)= cx - d

OpenStudy (israkjahim):

thanks you so much phi

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