Answered

Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.

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]

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.