[amp_mcq option1=”1″ option2=”3″ option3=”4″ option4=”9″ correct=”option1″]<\/p>\n
The maximum value of $y=x^2-6x+9$ obtained when $x$ varies over the interval $2$ to $5$ is $9$.<\/p>\n
To find the maximum value of a function, we can find the critical points and then compare the function’s value at the critical points and the endpoints of the interval.<\/p>\n
The derivative of $y=x^2-6x+9$ is $y’=2x-6$. The critical points of $y$ are the points where $y’=0$. Solving $y’=0$, we find that the critical point is $x=3$.<\/p>\n
Since $y’$ is a polynomial, it is defined for all real numbers. Therefore, the critical point $x=3$ is within the interval $2$ to $5$.<\/p>\n
To compare the function’s value at the critical point and the endpoints of the interval, we can evaluate $y$ at $x=2$, $x=3$, $x=4$, and $x=5$.<\/p>\n
We find that $y(2)=1$, $y(3)=9$, $y(4)=7$, and $y(5)=1$.<\/p>\n
Therefore, the maximum value of $y$ obtained when $x$ varies over the interval $2$ to $5$ is $9$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”1″ option2=”3″ option3=”4″ option4=”9″ correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20211","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n