## 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 $$[-1,~1]$$

1. IrishBoy123

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2. myininaya

hey @rational so we want the max D to be an element of [-1,1]?

3. myininaya

sorry distance

4. myininaya

not only that but we want it to be a minimum of [-1,1]?

5. rational

sorry $$x~\in ~[-1, 1]$$ we want to find $$m,~n$$ such that the deviation of $$f(x)$$ from $$x$$ axis is minimum

6. myininaya

oh okay

7. rational

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8. 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

9. 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

10. rational

yes there is no largest "maximum deviation" but there certainly exists a least "maximum deviation"

11. 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

12. 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.

13. 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 $$(m,n)=(0,-\frac{1}{2})$$

14. 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.

15. 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