Find the absolute maximum and minimum values of a functions given by f(x) =2x^3 - 15x^2+36x+1 on the interval [ 1,5].
Solution=> we have
F(x) =2x^3 - 15x^2 +36x+1
F'(x) =6x^2 - 30x+30 =6(x-3) (x-2)
Note tha f'(x) =0 gives x=2 and x=3
We shall now evaluate the value of F at these points and at end points of the interval [1,5] at x=1, x=2, x=3 and x=3 so
F(1) = 2(1^3 )-15(1^2 )+36(1)+1=24
F(2) = 2(2^3 )-15(2^2)+36(2)+1=29
F(3) = 2(3^3)-15(3^2)+36(3)+1=28
F(5) = 2(5^3 )-15(5^2 )+36(5)+1=56
Thus, we conclude that absolute maximum value of F on [1,5] is 56, occurring at x=5,and absolute minimum value of on [1,5] is 24 which occurs at x=1
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