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

Sagot :

konrad509
[tex]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=\dfrac{5}{11}}[/tex]
dln
Critical points are when the derivative is equal to zero
[tex]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 } [/tex]

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