The function f(x)=(x^4)-(10x^3)+(18x^2)-8 is continuous on the closed interval (1,8). Find the absolute minimum and maximum values for the function on this interval. Please help me!!! And please show your work so that i understand!! Thank you!!

Answer 1

find the derivative
dy/dx = 4x^3 - 30x^2 + 36x
set that equal to zero
4x^3 - 30x^2 + 36x= 0
2x(2x^2 - 15x + 18) = 0
2x(x-6)(2x-3) = 0
x = 0 or x = 6 or x = 3/2 , but x=0 lies outside our interval

so f(6) = ....
f(3/2) = ...
and for the end-values
f(1) = ...
f(8) = ...

you do the arithmetic, and decide which is the highest and lowest f(x) value.