- #1

Larrytsai

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f(x)= x+1 / x

attempt: x= y + 1 / y

multiply both sides by 'y'

x(y)= y + 1

subtract both sides by y

x(y) -y=1

divide 'x'

im lost from here help please:D

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- Thread starter Larrytsai
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- #1

Larrytsai

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f(x)= x+1 / x

attempt: x= y + 1 / y

multiply both sides by 'y'

x(y)= y + 1

subtract both sides by y

x(y) -y=1

divide 'x'

im lost from here help please:D

- #2

symbolipoint

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find an inverse such as x = (y + 1)/x and the rest is simple algebraic steps.

- #3

Larrytsai

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find an inverse such as x = (y + 1)/x and the rest is simple algebraic steps.

umm my book saids y= x+1/x

- #4

symbolipoint

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to x = y + 1/y

I made a variable writing error in my first response.

- #5

Larrytsai

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to x = y + 1/y

I made a variable writing error in my first response.

yea that's i did but what i don't know is where i went wrong in my algebra

- #6

symbolipoint

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... just to continue, your next step would be to multiply both sides by 'y'.

- #7

Larrytsai

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... just to continue, your next step would be to multiply both sides by 'y'.

k this is what i got so far x(y)= y-1

should i move the y to the other side or the '-1'?

- #8

symbolipoint

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k this is what i got so far x(y)= y-1

should i move the y to the other side or the '-1'?

No. That step is wrong. I obtained an xy term like you did, but I see no way to obtain a clear y as a function of x. Either I have become deficient in some of my inverse function skills, or your original function cannot be converted to an inverse according to "Intermediate Algebra" methods. Certainly someone will advise us. Maybe a different coordinate system? Polar?

- #9

HallsofIvy

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k this is a function and i have to find the inversed

f(x)= x+1 / x

attempt: x= y + 1 / y

multiply both sides by 'y'

x(y)= y + 1

First factor out "y": y(x- 1)= 1 and then divide by x-1.subtract both sides by y

x(y) -y=1

divide 'x'

However, if, as you appear to be saying, the problem really is y= x+ 1/x, becoming x= y+ 1/y, then multiplying both sides by y gives you xy= yim lost from here help please:D

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