Asked by Josh
Inverse functions
what is the inverse function of
f(x)=1/square root of x +1?
and f(x)=1-x^2, x<0?
what is the inverse function of
f(x)=1/square root of x +1?
and f(x)=1-x^2, x<0?
Answers
Answered by
Jai
f(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~ :)
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~ :)