Urgent, Please help! For f(x)=1+x+x2. Use the Mean Value Theorem to show that there is?
For f(x)=1+x+x2. Use the Mean Value Theorem to show that there is a real number c, 0<c<2, such that f(2)−f(0)=f′(c)(2−0).
For f(x)=1+x+x2. Use the Mean Value Theorem to show that there is a real number c, 0<c<2, such that f(2)−f(0)=f′(c)(2−0).
az_lender
Favorite Answer
Your teacher is a moron. The Mean Value Theorem ASSERTS that there is a real number c, 0<c<2, such that f(2) - f(0) = f'(c) * (2 - 0). No work is needed, if you are allowed to "use" the Mean Value Theorem.
But probably what your (moron, idiot) teacher is looking for is the actual number c that demonstrates the truth of the theorem when f(x) = 1 + x + x^2.
In this case, f'(x) = 1 + 2x, and you want a number c such that
(1 + 2 + 2^2) - (1 + 0 + 0^2) = (1 + 2c) * (2-0), or
7 - 1 = 2*(1 + 2c), or
3 = 1 + 2c,
implying that c = 1.
This illustrates the MVT, because 1 does in fact lie in the interval 0 < c < 2.
Never mind even trying to explain to your imbecile teacher that the question is phrased inappropriately. You could TRY pointing out that if he/she wanted the actual number c, he/she should have said "FIND" the number c. Etc etc. But I'm guessing your a.h. teacher is not bright enough to understand why that would have been the right way to phrase the question. Sorry you have to put up with him or her.
?
f(x) = 1 + x + x²
f(2) = 1 + 2 + 2² = 7
f(0) = 1 + 0 + 0² = 1
f(2) - f(0) 7 - 1
f '(c) = ------------- = ------ = 3
2 - 0 2
and
f(2) - f(0) = 7 - 1 = 6 = 3(2-0) = f '(c)(2 - 0)
so there exists some number c, 0 < c < 2, such that
f(2) - f(0) = 7 - 1 = 6 = 3(2-0) = f '(c)(2 - 0)