- #1

- 3

- 0

x, if x<1

x², if 1<x<9 (read x <or equal to 1 and x< or equal to 9)

27*sqrt(x), if x>9

I know that f is not differentiable at x=1 and x=9 because the derivative at each one of these points don't exist once the lateral limits don't coincide.(Is that right?)

Then, I know that f is differentiable at R-(1)-(9).

The inverse function exists where f exists and is continuous.The function is continuous in R and exists when x>0, right?So the inverse function exists when x belongs to R and is positive.

Now I need to know where the inverse function is differentiable.It will be differentiable where f is differentiable but when f ' (f⁻1(x)) is not 0?Considering that the inverse function is a mirror of the original function it will be continuous where f is continuous.Also it will exists where f exists and f ' (f⁻1(x)) is not 0.I don't know how to determine the points where the inverse function is differentiable and calculate the derivative of the inverse funtion at these points.I mean, I know how to calculate the derivative of the inverse function but I don't know where (which points) since I'm told to " calculate de derivative of the inverse function at the points where the inverse function is differentiable.

Also, why I only calculate the derivative of the function at the points 1 and 9 to determine the points where the function is differentiable?What assures me that there isn't another point that belongs to the domain where the function is not differentiable?