Question

Find the maximum and minimum of:
Y=x³-3x+5

Answer 1

$y=x^3-3x+5\\y'=3x^2-3\\\\3x^2-3=0\\3(x^2-1)=0\\3(x-1)(x+1)=0\\x_0=-1 \vee x_0=1$

If $x\in(-\infty,-1)\cup(1,\infty)$ then $y'>0$

If $x\in(-1,1)$ then $y'<0$

Therefore, there's a minimum at $x=1$ and a maximum at $x=-1$.

$y_{min}=1^3-3\cdot 1+5=1-3+5=3\\y_{max}=(-1)^3-3\cdot(-1)+5=-1+3+5=7$