Derivative of inverse function?

[thirteenheath]

Consider the function f: R-->R where F(x)=

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?

 

[SteveL27]

Consider the function f: R-->R where F(x)=

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?)

The derivative at x = 1 of f(x) = x is 1; and the derivative of f(x) = x² at x = 1 is 2. If you think of the tangent line traveling up the graph of this function, it has slope 1 all the way up to x = 1, then the slope suddenly becomes 2. That's the mental picture.

Mathematically, when you try to form the difference quotient to find the derivative of your function, the left-sided limit will be 1; and the right-sided limit (of the difference quotient) will be 2.

Since the left and right limits both exist but are not equal, by definition the limit of the difference quotient does not exist.

That's the reason f is not differentiable at x = 1.

Your intuition is correct but it's helpful to understand the precise technical reason why f isn't differentiable at x = 1. It's because the difference quotient has left- and right-sided limits that aren't equal to each other.

 

[LCKurtz]

Consider the function f: R-->R where F(x)=

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

...

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?

Because you know polynomials and $\sqrt x$ are differentiable on those intervals. It is only where the formula changes you potentially have problems.

 

[Mark44]

Consider the function f: R-->R where F(x)=

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

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