The given function is
[tex] f(x) = 4x^3-18x^2-48x+1 [/tex]
We need to find the derivative.And the derivative is
[tex] f'(x) = 4(3)x^2 -18(2)x-48 = 12x^2 -36x-48 [/tex]
Now we need to find the critical numbers, and for that we need to set f'(x) to 0 and solve for x, that is
[tex] 12x^2-36x-48=0
\\
12(x^2 -3x-4)=0
\\
x^2-3x-4=0
\\
(x-4)(x+1)=0
\\
x-4=0 , x+1=0
\\
x=4, x=-1
\\
x=-1,4 [/tex]
Now we need to find the value of the given functions at the boundary numbers and the critical numbers.
[tex] f(-2)=4(-8)-18(4)-48(-2)+1=-7
\\
f(-1) = 4(-1)-18(1)-48(-1)+1=27
\\f(4) = 4(64)-18(16)-48(4)+1=-223
\\
f(5)= 5(125)-18(25)-48(5)+1 =-64
[/tex]
So absolute minimum is -223 at x=4 and absolute maximum is 27 at x=-1 .
