find the maximum value of f(x) = x^3 - 3x^2 - 9x - 7 Not only the answer but all the development. Thank you
You need to remember that the function reaches its maximum at a value of x that cancels first derivative. Hence, you need to find the first derivative such that:
f'(x) =`3x^2 - 6x - 9`
If f'(x) = 0=> `3x^2 - 6x - 9` = 0 => `x^2 - 2x - 3 ` = 0
You need to find the roots of derivative such that:
`x^2 - 2x - 2 - 1` = 0 (notice the difference -2-1 substitutes -3)
Grouping the terms yields:
`(x^2 - 1) - (2x + 2) = 0`
`` Notice that the difference of squares `x^2 - 1` may be substituted by special product (x-1)(x+1).
Notice that you need also to factor out 2 such that:
(x-1)(x+1) - 2(x + 1) = 0
Factoring out (x+1) yields:
(x+1)(x -1 - 2) = 0
(x + 1)(x - 3) = 0 => x+ 1 = 0 => x = -1
x - 3 = 0 => x = 3
You need to decide what value of x denotes the maximum reached by f(x).
You need to select a value of x, larger than -1 and smaller than 3, hence x = 0.
Plug x = 0 in the equation of function such that:
f(0) = -7
The derivative has positive values over `(-oo,-1)U(3,oo).`
`` Hence, the function increases over`(-oo,-1), ` then decreases over (-1,3), hence the function reaches its maximum at x = -1.
You need to evaluate the maximum value of f(x) such that:
f(-1) = -1 - 3 + 9 - 7=-2
Hence, the function reaches its maximum at x = -1 and its value is f(-1)=-2.