Question

X^5(1-x)^6
How to find the critical points

Answer 1

`y=x^5(1-x)^6\\ y'=5x^4(1-x)^6+x^5\cdot6(1-x)^5\cdot(-1)\\ y'=x^4(1-x)^5(5(1-x)-6x)\\ y'=x^4(1-x)^5(5-5x-6x)\\ y'=x^4(1-x)^5(5-11x)\\\\ x^4(1-x)^5(5-11x)=0\\ \boxed{x=0 \vee x=1 \vee x=(5)/(11)}`

Answer 2

Critical points are when the derivative is equal to zero

`F(x)=x^( 5 )(1-x)^( 6 )\\ F'(x)=x^( 5 )*-6(1-x)^( 5 )+5x^( 4 )(1-x)^( 6 )\\ F'(x)=-6x^( 5 )(1-x)^( 5 )+5x^( 4 )(1-x)^( 6 )\\ -6x^( 5 )(1-x)^( 5 )+5x^( 4 )(1-x)^( 6 )=0\\ (x^( 4 )(1-x)^( 5 ))(-6x+5(1-x))=0\\ (x^( 4 )(1-x)^( 5 ))(5-11x)=0\\ \\ (x^( 4 )(1-x)^( 5 ))=0\\ x=0\ and\ 1\\ \\ (5-11x)=0\\ x=\frac { 5 }{ 11 } \\ \\ \boxed { Critical\ Points\ at\ x=\frac { 5 }{ 11 }\ ,0,\ and\ 1 }`