anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
phi
  • phi
ooh. Time to use the equation editor.
anonymous
  • anonymous
|dw:1443790834332:dw|
anonymous
  • anonymous
@phi finished

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phi
  • phi
the first part f( g(x) ) means everywhere you see x in f(x), replace it with g(x)
anonymous
  • anonymous
how exactly do I replace it.
imqwerty
  • imqwerty
Metal (:
phi
  • phi
the first step is \[ f( g(x) ) = \frac{g(x) -7}{g(x)+3} \]
phi
  • phi
but g(x) is also the messy \( \frac{-3x-7}{x-1} \) so we put that expression in for g(x)
phi
  • phi
btw, for future reference metal (iron or copper) is different from medal (shiny doodad)
phi
  • 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
  • anonymous
(-3x-7)-7 --------- (x-1)+3
anonymous
  • anonymous
?
phi
  • phi
\[ \frac{\frac{-3x-7}{x-1}-7}{\frac{-3x-7}{x-1}+3} \]
phi
  • phi
that is how you replace g(x) with the messy fraction
anonymous
  • anonymous
what happens next
phi
  • 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
  • anonymous
ok so would that cancel out x-1 or...
phi
  • phi
do the top first: distribute the (x-1) (which means multiply both terms inside by (x-1)
phi
  • 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
  • phi
can you simplify the top ?
anonymous
  • anonymous
yes, -3x ----- (x-1) x-1 I'm not sure
phi
  • phi
\[ \frac{-3x-7}{(x-1)}(x-1) \] that is the first term, the (x-1) cancels
anonymous
  • anonymous
if the bottom cancels then the top is -3x-7
phi
  • phi
yes. and the 2nd term is -7(x-1) if you distribute the -7 what do you get ?
anonymous
  • anonymous
-3x-7= -7x+7
phi
  • phi
there is no = sign, just -3x-7+ -7x + 7 (that mess simplified to that) now combine like terms
anonymous
  • anonymous
-10x
phi
  • 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
  • anonymous
you said the bottom right?
phi
  • phi
the (x-1) times the first term will cancel the (x-1) in the botom
phi
  • 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
  • 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
  • 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
  • anonymous
x-1 cancels out so 3-7 is 4 so -3-4.
anonymous
  • anonymous
-3x-4
phi
  • phi
x-1 cancels so you are left with -3x-7 (for the first term)
phi
  • phi
\[ \frac{-3x-7}{\cancel{(x-1)}}\cancel{(x-1)}+3(x-1) \\ -3x-7 +3(x-1) \]
phi
  • phi
now distribute the 3 in the 2nd term
anonymous
  • anonymous
-3x-7+3x-3
anonymous
  • anonymous
-10
phi
  • 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
  • phi
* f( g(x) ) =
anonymous
  • anonymous
1?
phi
  • 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
  • anonymous
1x
phi
  • 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
  • anonymous
is this right?
phi
  • phi
it's almost right, g(x) is (-3x-7)/(x-1)
phi
  • 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
  • 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
  • phi
follow?
anonymous
  • anonymous
um not really
phi
  • phi
do you follow this part \[ g(x) = \frac{-3x-7}{x-1} \\ g( f(x)) = \frac{-3f(x)-7}{f(x) -1} \]
anonymous
  • anonymous
yes
phi
  • 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
  • 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
  • anonymous
oh yea cancel out x-3 right
anonymous
  • anonymous
x+3
phi
  • 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
  • anonymous
yes I see that now.
phi
  • phi
now you see the (x+3) that would be nice to get rid of. multiply top and bottom by (x+3)
anonymous
  • anonymous
yes
phi
  • 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
  • phi
can you do the top ? post your steps. the first step is "distribute" the (x+3)
anonymous
  • anonymous
I thought x+3 cancels out?
phi
  • 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
  • anonymous
-3(x-7)-7 -3x+21-7 -3x+14 ?
phi
  • 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
  • phi
you did the first part correctly. -3(x-7) but you also have -7(x+3)
anonymous
  • anonymous
-3x+14-7x-21 4x-7
anonymous
  • anonymous
-10x-7
phi
  • phi
-3(x-7) + -7(x+3) try again
anonymous
  • anonymous
-3x-7+-7x+21 -10x+14
phi
  • phi
ok but how are you doing -3(x-7) ?
anonymous
  • anonymous
ugh I messed up again. -3x+21+-7x-21
anonymous
  • anonymous
-10x
phi
  • phi
yes, that looks good. now we do the bottom \[ \left(\frac{x-7}{x+3} -1\right)(x+3) \]
phi
  • phi
any luck?
anonymous
  • anonymous
no luck :(
phi
  • phi
can you distribute (x+3) by writing (x+3) next to each term inside the parens?
anonymous
  • anonymous
(x-7) -1(x+3)
phi
  • phi
ok, now distribute the -1
anonymous
  • anonymous
x-7-1x-3
anonymous
  • anonymous
-10
phi
  • phi
ok, so we have \[ g( f(x) ) = \frac{-10x}{-10} \]
anonymous
  • anonymous
x
phi
  • 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
  • anonymous
Thank you soo much :) can we do one more problem it won't take that long.
phi
  • phi
if you go fast
anonymous
  • 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
  • anonymous
options^
phi
  • phi
divide both sides by cos theta
anonymous
  • anonymous
45 degs
phi
  • phi
yes, you get sin x/cos x = 1 tan x = 1 x= atan 1 = 45º
phi
  • phi
or you can remember that sin 45 = sqr(2)/2 and cos 45 = the same thing

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