Fabien M. answered • 10/18/20
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f(x)=x^2−4x+9, [0,4]
when, x = 0
f(x) = x^2 -4x +9
f(0) = y = 0 - 0 + 9 = 9
when, x=4
f(5) = y = 16 - 16+9 = 9
thus, we have 2 points (0, 9) ; (4, 9)
slope,m = {9-(9)} / {4-0} = 0
hence, we have to calculate all the points,x where 0<x<9 and slope=0
f '(x) = 2x - 4 = 0
or, f '(c) = 2c - 4 = 0
c = 4/2 =2 ( 0<x<4)
hence, the there is only one solution c=2 which satisfies Rolle's theorem.
Fabien M.
tutor
Rolle's theorem is the result of the mean value theorem where under the conditions:
f(x) be a continuous functions on the interval [a, b]
and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a).
Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0.
In other words, under the above three conditions we can always find a tangent to the curve of f that is horizontal (slope = f'(c) = 0).
Rolle's Theorem
If f(x) is
1) a continuous function on the interval [a, b]
2) differentiable on the open interval (a, b)
3) and f(a) = f(b),
then there is at least one value c of x in the interval (a, b) such that
f '(c) = 0
Applying Rolle's theorem,
Main equation: f(x)=x^2−4x+9, interval: [0,4]
base on the theorem a =0, b=4
f(a) = f(b) is equivalent to f(0)=f(4) which is in our case, f(0)=f(4)=9
then f'(x)=0, which is in our case f'(x)= 2x - 4 = 0 equivalent to f'(c)=2c-4=0
so 2c-4=0 --> c = 4/2 =2. so the value for c is 2 and 2 is within the interval [0,4]. so it satisfies rolle's theorems.
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10/18/20
Precious W.
I do not understand the first two answers. can you explain more, please?10/18/20