Answer to Question #276508 in Calculus for RIYAD

Question

Evaluate maximum and minimum value of the function

ƒ(𝑥) =  𝑥³-3𝑥² + 3𝑥+1

Answer 1

Since function is unbounded, it can reach maximum or minimum only in extreme points

Lets find out extreme point of the function

$f'(x)=3x^2-6x+3\implies 3x^2-6x+3=0\implies x^2-2x+1=0\implies (x-1)^2=0\implies x-1=0\implies x=1$

x = 1 is a critical point of the function

f'(x) > 0 when x < 1 and f'(x) > 0 when x > 1, so the first derivative doesn't change its sign(the function is increasing for x < 1 and x > 1), so x = 1 is not an extreme point of the function, which means function has not any maximum or minimum points