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August 29, 2014

August 29, 2014

Posted by **Josh** on Sunday, October 23, 2011 at 1:38am.

what is the inverse function of

f(x)=1/square root of x +1?

and f(x)=1-x^2, x<0?

- Pre Calculus -
**Jai**, Sunday, October 23, 2011 at 2:09amf(x) = 1/sqrt(x+1)

we can rewrite this as

y = 1/sqrt(x+1)

to get the inverse function, we replace x by y, and y by x, and solve for y:

x = 1/sqrt(y+1)

sqrt(y+1) = 1/x

y + 1 = 1 / x^2

y = (1/(x^2)) - 1

f'(x) = (1/(x^2)) - 1

for #2, we do the same:

f(x) = 1 - x^2

y = 1 - x^2

x = 1 - y^2

y^2 = 1 - x

y = sqrt(1-x)

f'(x) = sqrt(1-x)

hope this helps~ :)

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