zeesbrat3 one year ago If f(x) is differentiable for the closed interval [−3, 2] such that f(−3) = 4 and f(2) = 4, then there exists a value c, −3 < c < 2 such that

1. zeesbrat3

@amoodarya

2. amoodarya

|dw:1436665630994:dw|

3. amoodarya

|dw:1436665699827:dw| every one can be f(x) so ...

4. amoodarya

|dw:1436665754136:dw| even this one

5. amoodarya

$f'(c)=\frac{f(b)-f(a)}{b-a}\\=\frac{4-4}{2-(-3)}=0$

6. amoodarya

|dw:1436665846936:dw|

7. zeesbrat3

How did you know it was f'(c)?

8. anonymous

Looks like they're testing you on Rolle's Theorem

9. anonymous

But you can also appeal to the Mean Value Theorem