Rolles Theorem and Mean Value Theorem.
1) Determine if Rolles Theorem applies to the given function on [0,3]. If so, find all numbers c on the interval that satisfy the theorem.
X^3 - 9x
Uh ok. So it has to be f ' (c) = 0 or something? Plug in random values from the interval and see if the derivative is gonna equal 0 ?
2) Determine if the given function satisfies the mean value theorem on [0,1]. If so, find all numbers c on the interval that satisfy the theorem.
6* squrt of(1-x^2)
No idea how to do this. Anybody?
1) Determine if Rolles Theorem applies to the given function on [0,3]. If so, find all numbers c on the interval that satisfy the theorem.
X^3 - 9x
Uh ok. So it has to be f ' (c) = 0 or something? Plug in random values from the interval and see if the derivative is gonna equal 0 ?
2) Determine if the given function satisfies the mean value theorem on [0,1]. If so, find all numbers c on the interval that satisfy the theorem.
6* squrt of(1-x^2)
No idea how to do this. Anybody?
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Rolles theorem states:
Let f(x) be continuous on [a,b] and differentiable on (a,b) then there exist a point such that f'(c) = ∆y/∆x = 0
f(3) = 27-27 = 0
f(0) = 0
So we can try finding a point.
The reason why we needed f(3) and f(0) is because we needed the slope of the secant line
f' = 3x^2 - 9 = 0
3x^2 = 9
x^2 = 3
x = ±√3
x= +√3
-√3 is not in domain so we dont care about it.
2.
f(0) = 6
f(1) = 0
(6-0) /(0-1) = -6
f' = 3(1-x^2)^-1/2 * -2x = -6x / √(1-x^2) = 0
x≠ 1
0 = -6x
x = 0
Let f(x) be continuous on [a,b] and differentiable on (a,b) then there exist a point such that f'(c) = ∆y/∆x = 0
f(3) = 27-27 = 0
f(0) = 0
So we can try finding a point.
The reason why we needed f(3) and f(0) is because we needed the slope of the secant line
f' = 3x^2 - 9 = 0
3x^2 = 9
x^2 = 3
x = ±√3
x= +√3
-√3 is not in domain so we dont care about it.
2.
f(0) = 6
f(1) = 0
(6-0) /(0-1) = -6
f' = 3(1-x^2)^-1/2 * -2x = -6x / √(1-x^2) = 0
x≠ 1
0 = -6x
x = 0