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Offlinesubiedude
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Registered: 02/06/09
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another math problem in need of help
    #11386334 - 11/04/09 08:20 PM (8 years, 8 months ago)

EXPONENTIAL AND LOGARITHMIC FUNCTIONS (INVERSE FUNCTIONS)



a) determine if the function is one-to-one
b) if the function is one-to-one, find a formula for the inverse.


f(x)= (x+5)^3


i know it's one-to-one, it's the formula im having trouble with. the cube is throwing me off.:mushroom2:


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InvisibleEntropymancer
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Re: another math problem in need of help [Re: subiedude]
    #11386354 - 11/04/09 08:21 PM (8 years, 8 months ago)

Howzabout x equals cube-root-of-y minus 5?


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Offlinekanglow
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Re: another math problem in need of help [Re: subiedude]
    #11386355 - 11/04/09 08:22 PM (8 years, 8 months ago)

Quote:

subiedude said:
the cube is throwing me off.:mushroom2:




then stop eating them before you math


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OfflineNeuron
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Re: another math problem in need of help [Re: kanglow]
    #11386362 - 11/04/09 08:23 PM (8 years, 8 months ago)

MATH LEARNING CENTER


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InvisibleEntropymancer
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Re: another math problem in need of help [Re: Neuron]
    #11386372 - 11/04/09 08:25 PM (8 years, 8 months ago)

Quote:

Neuron said:
MATH LEARNING CENTER



Blasphemy


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Offlinesubiedude
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Re: another math problem in need of help [Re: Entropymancer]
    #11386374 - 11/04/09 08:25 PM (8 years, 8 months ago)

Quote:

Entropymancer said:
Howzabout x equals cube-root-of-y minus 5?





WINNER:thumbup:


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Invisibledr_gonz

Registered: 08/18/03
Posts: 44,645
Re: another math problem in need of help [Re: subiedude]
    #11386396 - 11/04/09 08:28 PM (8 years, 8 months ago)

explain steps


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InvisibleEntropymancer
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Re: another math problem in need of help [Re: subiedude]
    #11386400 - 11/04/09 08:29 PM (8 years, 8 months ago)

This sort of algebra is really easy to think through.

You know you want to get rid of the exponent around (x+5) so you can get x by itself.  So just take each side to the 1/3rd power (which is to say, take the cube root of each side).  Then all you have to do is subtract the 5 of both sides, and :voila:


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OfflineTedwilto
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Re: another math problem in need of help [Re: Entropymancer]
    #11386468 - 11/04/09 08:37 PM (8 years, 8 months ago)

y=(x+5)^3
cube rootY = x+5
CuberootY - 5 = X
switch variables:

cuberootX - 5 = Y

I think thats the inverse unless im finding something else. who knows,


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Invisiblepwnasaurus
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Re: another math problem in need of help [Re: Tedwilto]
    #11386980 - 11/04/09 09:38 PM (8 years, 8 months ago)

Quote:

Tedwilto said:
y=(x+5)^3
cube rootY = x+5
CuberootY - 5 = X
switch variables:

cuberootX - 5 = Y

I think thats the inverse unless im finding something else. who knows,



This looks good to me.


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