Question

May someone assist me with the question:
Find the point on the graph of f(x)=x^2 that is closest to the point (2, (1/2)). Thank you :)

Answer 1

I got to the point which I took the derivative of the equation I got, leaving me with:
0=4(x^3+x-1), in which, synthetic division doesn't work, thus, I desire not to use a calculator to find its roots. I could have done something wrong, along the way to getting my answer, however.

Answer 2

If you mean the vertex.

It's 0,0

Although I am probably wrong.

Answer 3

@voce , I don't, thanks though :).

Answer 4

From my first post, I desire to know if I messed up, where, if not, what can I do from here?

Answer 5

i think it should be
\[d^2=(x-2)^2+(x^2-\frac{1}{2})^2\] which is a quadratic

Answer 6

a point on the parabola is (x, x^2). your point is (2, 1/2)

the distance between them is \[\sqrt{\left( x-2 \right)^2+\left( x^2-\frac{ 1 }{ 2 } \right)^2}\]
minimize this by taking the derivative and setting equal to 0 and solve for x.

Answer 7

@satellite73 , indeed, that's what I got, then I distributed, raised the power (to eliminate the radical), then took the derivative.

Answer 8

what @pgpilot326 said, although i would ignore the square root

Answer 9

oh, and i was wrong of course, it is not a quadratic

Answer 10

it is \[x^4-4 x+\frac{17}{4}\] i think

Answer 11

^, you removed it a step before me, but, this changes not the result, which I got was:
\[x^4+2x^2-4x+4.25\]
In which, upon taking the derivative:\[4x^3+4x-4\rightarrow 4(x^3+x-1)\]

Answer 12

there is no \(2x^2\) in it

Answer 13

\[2(x-2)+4x\left( x^2-\frac{ 1 }{ 2} \right)=0\]

Answer 14

@satellite73 , how not?

Answer 15

you get \(x^2\) from \((x-2)^2\) and \(-x^2\) from \((x^2-\frac{1}{2})^2\)

Answer 16

Ohh, you took the derivative, and distributed not...yes?

Answer 17

no i did not take the derivative yet
just was saying that \[(x-2)^2+(x^2-\frac{1}{2})^2=x^4-4 x+\frac{17}{4}\]

Answer 18

\[4x^3-2x+2x-4=4(x^3-1)\Rightarrow x = 1\]
is the only real root

Answer 19

so (1, 1) is the point on the parabola y = x^2 closest to the point (2. 1/2)

Answer 20

then you see the derivative is what @pgpilot326 wrote
\[4(x^3-1)\] but the \(x^2\) term drops out in the initial calculation, before taking the derivative (or after, since it is not there to begin with)

Answer 21

in other words \(x^4+2x^2-4x+4.25\) is wrong, it should be \(x^4-4x+4.25\)

Answer 22

Mmm, yes, but how does is drop out? I distributed incorrectly?

Answer 23

yes

Answer 24

it's easier if you don't multiply until after you take the derivative. just use the chain rule

Answer 25

Wait, I see the negative.

Answer 26

\((x-2)^2=x^2-4x+4\)
\((x^2-\frac{1}{2})^2=x^4-x^2+\frac{1}{4}\)

Answer 27

or you could make this an exercise in multiplying polynomials

Answer 28

motha-fuking negative...dang. Icy, icy now. Wow, that TOTALLY MONKEY WRENCHED THE WHOLE OUTCOME. Damn. Thanks a lot, the both of you. and, lol @pgpilot326

Answer 29

but i understand it's difficult to let go sometimes... got to find the errors and put them to rest and relieve the mind

Answer 30

and yes... lol

Answer 31

damn minus sign...

Answer 32

Such a minor factor created SUCH a wrong result. Thanks so much. I wish I could metal the both of you for your contribution. Thanks a lot.

Answer 33

welcome to math

Answer 34

medal*

Answer 35

whew... i thought thtt is some new thing the kids are doing... metalling

Answer 36

or this is really an episode of scooby doo and those "damn mettling kids" thwarted the minus sign