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OpenStudy (anonymous):

The point B(a,b) is on the curve f(x)=x² such that B is the point which is closest to A(6,0). Calculate the value of a.

OpenStudy (anonymous):

a point on the curve \(y=x^2\) looks like \((x,x^2)\) and distance between \((x,x^2)\) and \((0,6)\) is by the distance formula \[\sqrt{(x-0)^2+(x^2-6)^2}\] easier to work with the square of the distance, so minimize \[x^2+(x^2-6)^2\]

OpenStudy (anonymous):

i am guessing this is calc, so multiply out to get \[d=x^4-11 x^2+36\] then maybe take the derivative to find the critical points good from there?

OpenStudy (anonymous):

Oh, shoot, I'm sorry the question said A(6,0) not (0,6).

OpenStudy (anonymous):

should be easy now since \[d'=4x^3-22x=2x(x^2-11)\] zeros are easy and they will give you a minimum

OpenStudy (anonymous):

But it's the same, right? Just swap the numbers in the distance formula.

OpenStudy (anonymous):

in that case it is even easier

OpenStudy (anonymous):

yeah it is the same, just swap

OpenStudy (anonymous):

\[d^2=(x-6)^2+x^4=x^4+x^2-12 x+36\]

OpenStudy (anonymous):

oh actually now it is not easier. you have to find the zeros of a cubic function. i liked the last one more

OpenStudy (anonymous):

we can do it a different way, but you get exactly the same equation to solve, so it will be no easier

OpenStudy (anonymous):

Well, I am allowed to use a calculator. So I think the cubic equation won't be a problem. :)

OpenStudy (anonymous):

well then we can go right to the answer http://www.wolframalpha.com/input/?i=%28x-6%29^2%2Bx^4

OpenStudy (anonymous):

So since local min at x=1.33, is that the answer?

OpenStudy (anonymous):

yeah looks like that

OpenStudy (anonymous):

Ok thanks! :)

OpenStudy (anonymous):

yw

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