Question

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 4x3 − 18x2 − 48x + 1, [−2, 5]

Answer 1

The given function is

$f(x) = 4x^3-18x^2-48x+1$

We need to find the derivative.And the derivative is

$f'(x) = 4(3)x^2 -18(2)x-48 = 12x^2 -36x-48$

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

$12x^2-36x-48=0 \\ 12(x^2 -3x-4)=0 \\ x^2-3x-4=0 \\ (x-4)(x+1)=0 \\ x-4=0 , x+1=0 \\ x=4, x=-1 \\ x=-1,4$

Now we need to find the value of the given functions at the boundary numbers and the critical numbers.

$f(-2)=4(-8)-18(4)-48(-2)+1=-7 \\ f(-1) = 4(-1)-18(1)-48(-1)+1=27 \\f(4) = 4(64)-18(16)-48(4)+1=-223 \\ f(5)= 5(125)-18(25)-48(5)+1 =-64$

So absolute minimum is -223 at x=4 and absolute maximum is 27 at x=-1 .