Prove that the expression
is an equivalent form of the mean-value theorem.
Find the value of in terms of and when:
For parts (a) and (b) keep fixed with and find the limit of as tends to 0.
Proof. If is continuous on and differentiable on , then by the mean-value theorem we have
Letting and for some (since ), we have
(This follows since from our definitions, is the distance from . Then, since is somewhere in the interval its value must be plus some portion of the distance to . This portion is then , which is how we know .) Substituting and and ,
Now for parts (a) and (b).
- If , we have , so,
- If , we have . So,
on letter b
when you divide everything by h²
how come does the last term equals to -x-h²/3?
why is h still squared since it was to the third power and h³/h² = h¹
It is a typo, and it should be -x-h/3.
Small nit: in part b, your solutions breaks when . When , your approach using
should work, since
Hey, I think there’s a typo in the first line. The solution breaks when… ? I fixed up the other latex problem (sorry, there’s no way to edit your comment once it’s posted, and no way to preview the latex… I’m working on it). Let me know if I introduced errors.
You have a typo in the quadratic formula. The discriminant should be $D = b^2 – 4ac$. So it should be $4h^3/3$ under the square root.