Can c_1 and c_2 Make F an Antiderivative of f?

[daniel_i_l]

Homework Statement

Let f:[0,2]->R be defined as:
if 0 =< x =< 1 then f(x) = 4(x^3)
if 1 < x =< 2 then x = x^2 + 2
Prove or disprove:
There exist c_1 , c_2 in R so that F:[0,2]-R defined as:
if 0 =< x =< 1 then f(x) = x^4 + c_1
if 1 < x =< 2 then x = (x^3)/3 + 2x + c_2

Homework Equations

The Attempt at a Solution

Now my question is, why don't any c_1,c_2 make F an AD of f?
Can any shed some light on this?
Thanks.

 

[pardesi]

definition of antiderivative

let $$f(x)$$ be a function which is differentiable in an interval C then there exists a g such that $$g(x)=f^{'}(x)$$ then f is called the antiderivative of g.

ur function is not differentiable at $$x=1$$

 

[Dick]

I suppose, except for any c_1,c_2, F is not differentiable at x=1.

 

[daniel_i_l]

Thanks for your answers! I just want to clear one thing up:
In this case for example, if f wasn't defined at x=1 then would F be an AD of f for all c_1,c_2?
Thanks.

 

[Dick]

Absolutely, if f only needs to be defined on [0,1)U(1,2].

 

[daniel_i_l]

Thanks for clearing that up!

 

[pardesi]

yes that's right it just depends on ur domain of definition