Will METAL!!
Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.
f(x) = the quantity x minus seven divided by the quantity x plus three. and g(x) = quantity negative three x minus seven divided by quantity x minus one.

- anonymous

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- phi

ooh. Time to use the equation editor.

- anonymous

|dw:1443790834332:dw|

- anonymous

@phi finished

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## More answers

- phi

the first part f( g(x) ) means everywhere you see x in f(x), replace it with g(x)

- anonymous

how exactly do I replace it.

- imqwerty

Metal (:

- phi

the first step is
\[ f( g(x) ) = \frac{g(x) -7}{g(x)+3} \]

- phi

but g(x) is also the messy \( \frac{-3x-7}{x-1} \)
so we put that expression in for g(x)

- phi

btw, for future reference metal (iron or copper) is different from medal (shiny doodad)

- phi

everywhere you see g(x) in
\[ \frac{g(x) -7}{g(x)+3} \]
erase the g(x) and replace it with
\[ \frac{-3x-7}{x-1} \]
you get a big mess, but it will simplify with some algebra

- anonymous

(-3x-7)-7
---------
(x-1)+3

- anonymous

?

- phi

\[ \frac{\frac{-3x-7}{x-1}-7}{\frac{-3x-7}{x-1}+3} \]

- phi

that is how you replace g(x) with the messy fraction

- anonymous

what happens next

- phi

to simplify , I would multiply top and bottom by (x-1)
\[ \frac{\left(\frac{-3x-7}{x-1}-7\right)}{\left(\frac{-3x-7}{x-1}+3\right)} \cdot \frac{(x-1)}{(x-1)}\]

- anonymous

ok so would that cancel out x-1 or...

- phi

do the top first:
distribute the (x-1) (which means multiply both terms inside by (x-1)

- phi

the top is
\[ \left(\frac{-3x-7}{x-1}-7\right)(x-1) \\ \frac{-3x-7}{(x-1)}(x-1) -7(x-1)
\]

- phi

can you simplify the top ?

- anonymous

yes, -3x
----- (x-1)
x-1
I'm not sure

- phi

\[ \frac{-3x-7}{(x-1)}(x-1) \]
that is the first term, the (x-1) cancels

- anonymous

if the bottom cancels then the top is -3x-7

- phi

yes.
and the 2nd term is
-7(x-1)
if you distribute the -7 what do you get ?

- anonymous

-3x-7= -7x+7

- phi

there is no = sign, just
-3x-7+ -7x + 7
(that mess simplified to that)
now combine like terms

- anonymous

-10x

- phi

yes the top (of the original fraction) simplifies to -10x
now we do the bottom
\[ \left(\frac{-3x-7}{x-1}+3\right)(x-1) \]

- anonymous

you said the bottom right?

- phi

the (x-1) times the first term will cancel the (x-1) in the botom

- phi

in case you lost track, we are doing the bottom of this mess:
\[ \frac{\left(\frac{-3x-7}{x-1}-7\right)}{\left(\frac{-3x-7}{x-1}+3\right)} \cdot \frac{(x-1)}{(x-1)} \]

- phi

we found that
\[ \frac{\left(\frac{-3x-7}{x-1}-7\right)}{\left(\frac{-3x-7}{x-1}+3\right)} \cdot \frac{(x-1)}{(x-1)} = \frac{-10x}{\left(\frac{-3x-7}{x-1}+3\right)(x-1)} \]
and we are doing the bottom
\[ \left(\frac{-3x-7}{x-1}+3\right)(x-1)\]

- phi

and the bottom
\[ \left(\frac{-3x-7}{x-1}+3\right)(x-1)\\ \frac{-3x-7}{(x-1)}(x-1)+3(x-1) \]

- anonymous

x-1 cancels out so 3-7 is 4 so -3-4.

- anonymous

-3x-4

- phi

x-1 cancels so you are left with -3x-7 (for the first term)

- phi

\[ \frac{-3x-7}{\cancel{(x-1)}}\cancel{(x-1)}+3(x-1) \\
-3x-7 +3(x-1) \]

- phi

now distribute the 3 in the 2nd term

- anonymous

-3x-7+3x-3

- anonymous

-10

- phi

ok, so the bottom simplifies to -10
we found that
\[ f(x) = \frac{\frac{-3x-7}{x-1}-7}{\frac{-3x-7}{x-1}+3} = \frac{-10x}{-10} \]
one more step to go

- phi

* f( g(x) ) =

- anonymous

1?

- phi

you can't ignore the x
remember if you have
\[ \frac{-10 \cdot x }{-10} \]
that is the same as
\[ \frac{-10}{-10} \cdot \frac{x}{1} \]

- anonymous

1x

- phi

yes or just x
you just showed f( g(x) ) = x
the question was
***Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x. ***
you just did the first part.
now they want to do it the "other way"
show
g( f(x) ) = x
I hope you are a "high-energy" person, because it's more work.

- anonymous

is this right?

- phi

it's almost right, g(x) is (-3x-7)/(x-1)

- phi

it can get confusing unless we are slow and methodical
\[ g(x) = \frac{-3x-7}{x-1} \\ g( f(x)) = \frac{-3f(x)-7}{f(x) -1} \]

- phi

now replace f(x) with its expression
\[ g( f(x)) = \frac{-3\frac{x-7}{x+3}-7}{\frac{x-7}{x+3} -1} \]

- phi

follow?

- anonymous

um not really

- phi

do you follow this part
\[ g(x) = \frac{-3x-7}{x-1} \\ g( f(x)) = \frac{-3f(x)-7}{f(x) -1} \]

- anonymous

yes

- phi

and f(x) is "short-hand" (or the name for)
\[ \frac{x-7}{x+3} \]
so everywhere we see f(x), we can put in the "long form"

- phi

\[ g( f(x)) = \frac{-3f(x)-7}{f(x) -1} \]
the top says "multiply -3 times f(x)" then subtract 7
in math, and replacing f(x) with its expression, we would write
-3 * (x-7)/(x+3) then subtract 7:
-3 * (x-7)/(x+3) -7
\[ -3 \cdot \frac{x-7}{x+3} - 7\]

- anonymous

oh yea cancel out x-3 right

- anonymous

x+3

- phi

we haven't gotten that far, but that will be the idea. First, do you see how we get
\[ g( f(x)) = \frac{-3\frac{x-7}{x+3}-7}{\frac{x-7}{x+3} -1} \]

- anonymous

yes I see that now.

- phi

now you see the (x+3) that would be nice to get rid of. multiply top and bottom by (x+3)

- anonymous

yes

- phi

\[ \frac{\left(-3\cdot \frac{x-7}{x+3}-7\right)(x+3)}{\left(\frac{x-7}{x+3} -1\right)(x+3)} \]

- phi

can you do the top ? post your steps. the first step is "distribute" the (x+3)

- anonymous

I thought x+3 cancels out?

- phi

distribute means multiply each term inside the parens
for example:
(a + b)(x+3) = a(x+3) + b(x+3)
use that same rule on
\[ \left(-3\cdot \frac{x-7}{x+3}-7\right)(x+3) \]

- anonymous

-3(x-7)-7
-3x+21-7
-3x+14 ?

- phi

"terms" are things multiplied together. the -3 * (x-7)/(x+3) is all one term
terms are separated by + or -. in other words, the -7 is also a term
Here is what it looks like, after distributing the (x+3)
\[ \left(-3\cdot \frac{x-7}{x+3}-7\right)(x+3)\\
-3\cdot \frac{x-7}{x+3}\cdot (x+3) -7 \cdot (x+3)
\]
notice the -7 is also multiplied by (x+3)

- phi

you did the first part correctly.
-3(x-7)
but you also have
-7(x+3)

- anonymous

-3x+14-7x-21
4x-7

- anonymous

-10x-7

- phi

-3(x-7) + -7(x+3)
try again

- anonymous

-3x-7+-7x+21
-10x+14

- phi

ok but how are you doing -3(x-7)
?

- anonymous

ugh I messed up again.
-3x+21+-7x-21

- anonymous

-10x

- phi

yes, that looks good.
now we do the bottom
\[ \left(\frac{x-7}{x+3} -1\right)(x+3) \]

- phi

any luck?

- anonymous

no luck :(

- phi

can you distribute (x+3)
by writing (x+3) next to each term inside the parens?

- anonymous

(x-7) -1(x+3)

- phi

ok, now distribute the -1

- anonymous

x-7-1x-3

- anonymous

-10

- phi

ok, so we have
\[ g( f(x) ) = \frac{-10x}{-10} \]

- anonymous

x

- phi

yes, so you have shown that
f(g(x))= x
and
g(f(x))= x
which proves that f(x) and g(x) are inverses.

- anonymous

Thank you soo much :)
can we do one more problem it won't take that long.

- phi

if you go fast

- anonymous

Find the angle θ (if it exists) in the interval [0°, 90°) for which sin θ = cos θ.
θ = 30°
θ = 45°
No such angle exists.
θ = 60°

- anonymous

options^

- phi

divide both sides by cos theta

- anonymous

45 degs

- phi

yes, you get
sin x/cos x = 1
tan x = 1
x= atan 1 = 45º

- phi

or you can remember that sin 45 = sqr(2)/2 and cos 45 = the same thing

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