You need to find the critical values of functon, hence, you need to solve the equation f'(x) = 0 such that:

`f'(x) = (x-5)^3 + 3(x-1)(x-5)^2`

You need to solve the equation f'(x) = 0 such that:

`(x-5)^3 + 3(x-1)(x-5)^2 = 0`

You need to factor out `(x-5)^2 ` such that:

`(x-5)^2(x - 5 + 3x - 3) =`  0

`(x-5)^2 = 0 =gt x_(1,2) = 5`

`4x - 8 = 0 =gt 4x = 8 =gt x_3 = 2`

a) You need to select the end values of interval [1,4] to evaluate if derivative is positive or negative at these values such that:

`x = 1 =gt f'(1) = (1-5)^2(4-8) = 16*(-4) = -64lt0`

`x=4 =gt f'(4) = (4-5)^2(16-8) = 1*8 =8gt0`

Hence, the function reaches its absolute minimum over interval [1,4] at x = 2.

b) Notice from a) that the function decreases over [1,2] and it increases over [2,4], hence, you need to select the a value from interval [4,8] to evaluate if derivative is positive or negative such that:

`x =6 =gt f'(6) = (6-5)^2(24-8) = 1*(16) = 16gt0`

`x = 4.5 =gt f'(4.5) = (4.5-5)^2(18-8) gt 0`

Hence, the function reaches its absolute minimum at x = 2 and it has no absolute maximum over [1,8].

c) Notice that the function has no absolute maximum or minimum over [4,9].