You need to remember that the critical values of a function express the zeroes of derivative of function, hence you need to find derivative of f(x).
`f'(x) = (x^3+ax^2+bx+c)' =gt f'(x) = 3x^2 + 2ax + b`
The problem provides you the following information: the function reaches a relative maximum at x = -1, hence f'(-1)=0
You need to substitute -1 for x in equation of f'(x) such that:
`f'(-1) = 3 - 2a + b =gt 3 - 2a + b = 0 =gt -2a + b = -3`
This equation is not sufficient to find a and b, hence you need one more information. The problem provides this additional information such that the function reaches a relative minimum at x = 3, hence f'(3)=0.
You need to substitute 3 for x in equation of f'(x) such that:
`f'(3) = 27 + 3a + b =gt 3a + b = -27`
Subtracting f'(-1) from f'(3) yields: `6a + b + 2a - b = -27 + 3`
Adding like terms yields: `8a= -24 =gt a = -24/8 =gt a = -3`
You need to substitute -3 for a in equation of f'(-1) such that: 6 + b = -3 => b = -9
Hence, evaluating the values of coefficients a and b under the given conditions yields a = -3; b = -9.