1. ## differentiable function

Question: True or False, Justify.
There does not exist a differentiable function f:$\displaystyle \Re$->$\displaystyle \Re$ such that f(-1)=9, f(3)=1 and f'(x)$\displaystyle \geq$-1 for all x $\displaystyle \in$ [-1,3].

I think this is false because in order to have a derivative of -1, you would have to have an original function as a line with a slope of -1, but there is only one line between these two points and that slope is -2.

What do you think?

2. ## Re: differentiable function

Originally Posted by CountingPenguins
Question: True or False, Justify.
There does not exist a differentiable function f:$\displaystyle \Re$->$\displaystyle \Re$ such that f(-1)=9, f(3)=1 and f'(x)$\displaystyle \geq$-1 for all x $\displaystyle \in$ [-1,3].

I think this is false because in order to have a derivative of -1, you would have to have an original function as a line with a slope of -1, but there is only one line between these two points and that slope is -2.

What do you think?

Use LaGrange theorem.

3. ## Re: differentiable function

Originally Posted by Also sprach Zarathustra
Use LaGrange theorem.
Most would call this the "Mean Value Theorem" (which is what should be used )

@Penguins, you are correct. No such function exists with the stated properties. Hope you can prove it!