Question

Someone please help?

Consider the function f(x)=x^2-4x+4 on the interval [0,4]. Verify that this function satisfies the three hypotheses of Rolle's Theorem on the interval.
f(x) is ?? on [0,4]
f(x) is ?? on (0,4)
and f(0)=f(4)=??

Then by Rolle's theorem, there exists a 'c' such that f'(c)=0.
Find the value of c.
c=??

Answer 1

continuous for 1st question, differentiable for 2nd question. For Rolle's Theorem, the endpoints "a" and "b" have to have f(a) = f(b) = 0, so you have to alter f(x) somewhat, because it is not satisfying Rolle's as is.

Answer 2

One such function would be your above function if you eliminated the "+4" at the end of the function.

Answer 3

So I guess the Rolle's Theorem states that if there is a second value that is the same as a first value then there has to be a zero slope some where in the function. The function you are givin is a quadratic so you can find the zero by finding the verxet x=-b/2a or taking the derivitive and solving for the zeros.

Answer 4

wouldn't the derivative be 2x-4 and to find the zeros I would just set it equal to zero.. then x would equal 2. ?

Answer 5

Once you alter f(x) to be x^2 - 4x, you can use Rolle's, so there is then some "x" where f ' (x) = 2x - 4 = 0 and that would be at x = 2. Have to have f(0) and f(4) = 0. Legitimate to alter original function since the new function will differ by only a constant, so your f '(x) of the new function will = the derivative of the original.

Answer 6

okay, c=2
but I'm still confused about the third part

Answer 7

Yes, you have the general idea. Just remembe that the endpoints have to have f(x) = 0. Some authors say that the f(x) at the endpoints can be just equal, but this conceptually resolves down to being equal to 0 with a translation of the graph.

Answer 8

If by the third part you are referring to f(0) and f(4) and what they equal, first off, they equal the same value. For the original equation, that value is "4". It is better to shift the graph down by 4 so that the new function has them both equal to 0.

Answer 9

how did you know it was 4 though?

Answer 10

ohhhh the "+4"

Answer 11

yes.

Answer 12

So, the function would equal 0 but because the +4 is there it's 4 and not 0.?

Answer 13

Yes, you are understanding this well. Apparently your author allows for f(a) = f(b) but NOT = 0. If f(a) does not equal f(b), then you enter a consideration of a more general theorem called the Mean Value Theorem.

Answer 14

Oh okay, that makes sense! (: thank you for clarifying!

Answer 15

you're welcome!