anonymous
  • anonymous
Simplify. f^-1g^-1 a. f/g b. g/f c. 1/fg d. fg
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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johnweldon1993
  • johnweldon1993
well f^-1 would be written as \[\frac{ 1 }{ f }\] g^-1 would be written as \[\frac{ 1 }{ g }\] so together you have \[\frac{ 1 }{ f } * \frac{ 1 }{ g }\] which is?
anonymous
  • anonymous
fg?
johnweldon1993
  • johnweldon1993
\[\frac{ 1 * 1 }{ f * g }\]

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More answers

Jhannybean
  • Jhannybean
f and g are two separate functions i believe
anonymous
  • anonymous
1/fg idk im so confused
Jhannybean
  • Jhannybean
what are you confused about?
johnweldon1993
  • johnweldon1993
1/fg would be what I get yes.....but @Jhannybean what'd you mean by that? lol
anonymous
  • anonymous
i dont understand how f^-1 went to 1/f and how g^-1 went to 1/g
johnweldon1993
  • johnweldon1993
ahh okay.....this is what happens when you want to turn a negative power ^-1 into a positive power ^1 what you do...is you put the function under a fraction under 1 a.k.a \[f ^{-1} = \frac{ 1 }{ f }\] *because no one likes working with negative exponents....so we turn them positive by doing this
anonymous
  • anonymous
oh ok i think i get it now ok so for example if i have y^-3 it would be 3/y ???
johnweldon1993
  • johnweldon1993
not quite.....the function would STILL be under 1.... \[y ^{-3} = \frac{ 1 }{ y^3 }\] the exponent follows the 'y'....but just becomes positive
johnweldon1993
  • johnweldon1993
hmmm....another way of looking at it....you know what a reciprocal is correct?
anonymous
  • anonymous
yes
johnweldon1993
  • johnweldon1993
perfect.....you know it is just the fraction flipped upside down \[\frac{ 1 }{ 2 } reciprocal = \frac{ 2 }{ 1 }\] like that well...when you have a function...raised to a negative power....a.k.a your \[f^{-1}\] look at the reciprocal of 'f' ...instead of 'f' it would be \[\frac{ 1 }{ f }\] and just bring the exponent with the 'f' \[f^{-1} = \frac{ 1 }{ f^1 } = \frac{ 1 }{ f }\] better?
anonymous
  • anonymous
a little better, yes
johnweldon1993
  • johnweldon1993
now let me test you \[f^{-5} = ?\]
anonymous
  • anonymous
1/f^5
johnweldon1993
  • johnweldon1993
perfect! :)
anonymous
  • anonymous
yay
johnweldon1993
  • johnweldon1993
so now back to the original question...do you see why the answer would be \[\frac{ 1 }{ fg }\]
anonymous
  • anonymous
because f^-1 is 1/f and g^-1 is 1/g
johnweldon1993
  • johnweldon1993
right....and then when you multiply them you get the 1/fg :)
anonymous
  • anonymous
thank u for your help :D
johnweldon1993
  • johnweldon1993
anytime! :)

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