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rational
 one year ago
Consider a quadratic function \(f(x)=x^2 + mx + n\).
Find the real numbers \(m\) and \(n\) such that the maximum distance between \((x,0)\) and \((x, f(x))\) is minimum in the interval \([1,~1]\)
rational
 one year ago
Consider a quadratic function \(f(x)=x^2 + mx + n\). Find the real numbers \(m\) and \(n\) such that the maximum distance between \((x,0)\) and \((x, f(x))\) is minimum in the interval \([1,~1]\)

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myininaya
 one year ago
Best ResponseYou've already chosen the best response.0hey @rational so we want the max D to be an element of [1,1]?

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0not only that but we want it to be a minimum of [1,1]?

rational
 one year ago
Best ResponseYou've already chosen the best response.1sorry \( x~\in ~[1, 1] \) we want to find \(m,~n\) such that the deviation of \(f(x)\) from \(x\) axis is minimum

rational
 one year ago
Best ResponseYou've already chosen the best response.1dw:1432795011006:dw

rational
 one year ago
Best ResponseYou've already chosen the best response.1we get different max deviations for different values of m,n need to find m,n that give minimum value for the max deviation

myininaya
 one year ago
Best ResponseYou've already chosen the best response.0I think I might be interpreting things wrong Take f(x)=x^2+n for example 0 is in [1,1] and the distance between (0,f(0)) and (0,0) is largest but the thing is there is no largest real number n

rational
 one year ago
Best ResponseYou've already chosen the best response.1yes there is no largest "maximum deviation" but there certainly exists a least "maximum deviation"

rational
 one year ago
Best ResponseYou've already chosen the best response.1for \(f(x,n) = x^2+n\), it seems the least "maximum deviation" will be less than \(1\) https://www.desmos.com/calculator/rtanhulgw9

Empty
 one year ago
Best ResponseYou've already chosen the best response.2My intuition tells me the correct answer is going to be \[f(x) = x^2\frac{1}{2}\] The maximum deviation of f(x) from the line y=0 is the absolute value of f(x), and f(x) at the end points to be 1/2 and vertex to be 1/2. Any other value of m will increase these values. Look at this graph of the absolute value. https://www.desmos.com/calculator/9cozemnmgj when m>0 the endpoints are larger than 1/2 and when m<0 the vertex is larger than 1/2, maknig m=0 the local min.

rational
 one year ago
Best ResponseYou've already chosen the best response.1that looks very convincing, so if i understand correctly the least "max deviation" for \(f(x)=x^2 + mx + n\) is \(\frac{1}{2}\) and is achieved when \((m,n)=(0,\frac{1}{2})\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, that's what I'm thinking, I wish I could come up with a better argument but it seems kinda obviously right to me. In my mind, it has to be at m=0 since it minimizes the derivative in a sense on the interval [1,1] and the reason that matters is because the derivative is what causes f(x) to increase.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Interestingly when I first did this I calculated the average minimum deviation and got m=0 and n=1/3 which is slightly different, but also useful and definitely practical in physics. =D
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