Calculus and the Absolute Extrema?

Question

I am having trouble with this problem, can anyone help?
Find the absolute extrema of the function f(x) = x^3 - 6x^2 + 9x - 8 on the interval [1,5]

Answer 1

Extrema occur where dy/dx = 0
3x^2 - 12x + 9 = 0
x^2 -4x + 3 = 0
(x -3)(x-1) = 0.
there are extremes (or inflection points at x= 3, and x = 1)

Plug them back in the original, and get your y values.

But !!!!
You also have to check the endpoints x = 1 and x= 5.

Note: a couple answers below are correct, but they forgot to check the endpoints.

Answer 2

f´(x) = 3x^2-12x+9
x^2-4x+3=0 x=((4+-sqrt(4))/2 x= 3 and x= 1
Caculate f(1) = -4 an extreme of the interval
f(3)=-8 absolute and local minimum
f(5) = 12 absolute maximum

Answer 3

f'(x) = 3x^2 - 12x + 9

the extrema are the largest & smallest values of the function in a given area. In this case, it would be where the derivitive (done above) equals 0 between 1 and 5

so 3x^2 - 12x + 9 = 0 = (3x - 9)(x - 1)

x = 1, 3 are the points.

Answer 4

f'(x) = 3x^2 - 12x + 9 =set= 0
x^2 - 4x + 3 = 0
(x-3)(x-1) = 0
Two points of extrema: x = 3,1
f(3) = 3^3 - 6(3)^2 + 9(3) - 8
= 9 - 6(9) + 27 - 8
= 9 - 54 + 27 - 8 = -26
f(1) = 3 - 6 + 9 - 8 = -2

Answer 5

Yes