We need to find a value of x in the interval [1,3]
such that f'(x) = 0. So we need to find f'(x)
f(x)= (x-1)(x-2)(x-3)
First we'll multiply out the right sides:
f(x)= (x²-3x+2)(x-3)
f(x)= x³-3x²-3x²+9x+2x-6
f(x)= x³-6x²+11x-6
Now we'll take the derivative:
f'(x) = 3x²-12x+11
Now we set the right side equal to zero in
hopes of getting a value of x in the interval [1,3]
3x²-12x+11 = 0
x =
x =
x =
x =
x =
x =
x =
x =
Using the +,
x =
≐ 2.57735 wjich is in the interval [1,3]
Using the -,
x =
≐ 1.42265 whis is also in the interval [1,3]
So there are two values of x in the interval [1,3] where f'(x) = 0.
[There only needs to be one such value to satisfy Rolle's theorem,
but it is just fine if there are more than one.]
Edwin