rational
  • rational
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 \([-1,~1]\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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IrishBoy123
  • IrishBoy123
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myininaya
  • myininaya
hey @rational so we want the max D to be an element of [-1,1]?
myininaya
  • myininaya
sorry distance

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myininaya
  • myininaya
not only that but we want it to be a minimum of [-1,1]?
rational
  • rational
sorry \( x~\in ~[-1, 1] \) we want to find \(m,~n\) such that the deviation of \(f(x)\) from \(x\) axis is minimum
myininaya
  • myininaya
oh okay
rational
  • rational
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rational
  • rational
we get different max deviations for different values of m,n need to find m,n that give minimum value for the max deviation
myininaya
  • myininaya
I 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
  • rational
yes there is no largest "maximum deviation" but there certainly exists a least "maximum deviation"
rational
  • rational
for \(f(x,n) = x^2+n\), it seems the least "maximum deviation" will be less than \(1\) https://www.desmos.com/calculator/rtanhulgw9
Empty
  • Empty
My 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
  • rational
that 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
  • Empty
Yeah, 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
  • Empty
Interestingly 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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