Friday

July 25, 2014

July 25, 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~ :)

**Related Questions**

Pre-Calculus - Find the inverse function for the function f(x)=2^(x)+1/2^(x)-1 ...

pre calculus - Find the inverse of the function below. Graph the function below ...

Calculus - F(x) = 1-x and g(x) = 1/x These functions have the property that f = ...

Algebra 2 - Ok, so I'm trying to find the inverse of each function. We're ...

math help1 - Okay I am a little bit confused about one math question can you ...

Calculus (Derivatives of Inverse Functions) - Suppose f(x) = sin(pi cos(x)). On ...

Pre-Calc - Does y=1/x have an inverse? It is a one-to-one function, so it should...

Math - Inverse Functions - Find the inverses of the following functions. y = 3(x...

Calculus-Inverse Functions - is the inverse function of x+ floor(x) equal to x...

Calculus - Integrate [1/square root of(e^(2x)-1)]. I have to use u substitution...