zeesbrat3
  • zeesbrat3
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
Mathematics
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zeesbrat3
  • zeesbrat3
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
Mathematics
chestercat
  • chestercat
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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zeesbrat3
  • zeesbrat3
amoodarya
  • amoodarya
|dw:1436665630994:dw|
amoodarya
  • amoodarya
|dw:1436665699827:dw| every one can be f(x) so ...

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amoodarya
  • amoodarya
|dw:1436665754136:dw| even this one
amoodarya
  • amoodarya
\[f'(c)=\frac{f(b)-f(a)}{b-a}\\=\frac{4-4}{2-(-3)}=0\]
amoodarya
  • amoodarya
|dw:1436665846936:dw|
zeesbrat3
  • zeesbrat3
How did you know it was f'(c)?
anonymous
  • anonymous
Looks like they're testing you on Rolle's Theorem
anonymous
  • anonymous
But you can also appeal to the Mean Value Theorem

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