Posted by **kobe** on Tuesday, December 1, 2009 at 7:29pm.

prove that the following function is differentiable at x=0 using first principles:

f(x)= e^x when x<0

x=1 when x>0 or x=0

also is f(x) differntiable for all real x?

## Answer This Question

## Related Questions

- Maths - Let f be a function such that f(−6)=−6, f(6)=6, f is ...
- Math - The twiceâ€“differentiable function f is defined for all real numbers and ...
- Calculus - Let f be a function such that f(−6)=−6, f(6)=6, f is ...
- math - If g is a differentiable function such that g(x) < 0 for all real ...
- calculus ap - If g is a differentiable function such that g(x) is less than 0 ...
- Maths Derivatives & Calulus - 1) Find d^2y / dx^2 given that y^3 - x^2 = 4 2)If ...
- calculus - a) Obtain all solutions of the equation z^3 +1 = 0 b) Let z = x + iy...
- Calculus - Assuming that f and g are functions differentiable at a (though we do...
- Math--Calculus - I'm having a tough time figuring out this problem... S(x) = ...
- Math - Is the step function differentiable in its domain? I answered it and said...

More Related Questions