Find the local maximum and minimum turning points of f. f= x^3=3x^2-9x+5
I assume that the 2nd '=' sign is supposed to be a '+' sign, and the function f() is then,
f(x) = x^3 + 3x^2 - 9x + 5
For turning points, f '(x) = 0 for some value(s) of x
Differentiating,
f '(x) = 3x^2 + 6x - 9 = 0
x^ + 2x - 3 = 0
factorising,
(x + 3)(x - 1) = 0
i.e. x = 1, and x = -3.
f(1) = 1 + 3 - 9 + 5 = 0
f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + 5 = -27 + 27 + 27 + 5 = 32
So, the turning points are: (1, 0), (-3, 32)