Question

Find the point on the graph f(x)=x^2 that is closest to the point (2,(1/2))

Answer 1

Using calculus

Answer 2

know distance formula ?

Answer 3

I feel like I should x)

Answer 4

I dont

Answer 5

Distance between points (x1,y1) and (x2,y2) is
\(\huge d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

Answer 6

now, one of the point is 2,1/2
can u find other point's co-ordinates in terms of x ??

Answer 7

u need to minimize this distance

Answer 8

to minimize the distance, u can also minimize the square of distance, that would be much more easier.

Answer 9

@hartnn how do make the equation that large?

Answer 10

type \large or \huge before the text
`\huge x=` will give you
\(\huge x=\)

Answer 11

@Fey , can u find other point's co-ordinates in terms of x ??

Answer 12

oh, so there's huge. i only know large. thnx

Answer 13

Oh, u mean. a^2+b^2=c^2?

Answer 14

not exactly that, but yes.

Answer 15

I get what u mean. Can u help me do it thhat way?

Answer 16

ok, we want a point on f(x)=x^2
let its x co-ordinate be 'x'
then its y co-ordinate = ?

Answer 17

y?

Answer 18

and y=f(x)=x^2
so the point is (x,x^2)
got this ?

Answer 19

ohh! Yeah, ok.

Answer 20

now u have 2 points, use distance formula to get the distance between them

Answer 21

(x,x^2) and (2,1/2)

Answer 22

ok. gimme a minute

Answer 23

ok, do I simplify it?

Answer 24

as i said, its lot easier to minimize the SQUARE of distance, so put the points in this formula
\(\huge d^2={(x_1-x_2)^2+(y_1-y_2)^2}\)
First put it and tell me what u get for d^2

Answer 25

oh, ok. \[d ^{2}=(x-2)^{2}+(x ^{2}-\frac{ 1 }{ 2})^{2}\]

Answer 26

absolutely! now to minimize d^2 , what u do ?

Answer 27

take the derivative?

Answer 28

yes! and equate it to ... ?

Answer 29

0!!! :D

Answer 30

yes :)
do it and tell me what u get...

Answer 31

is the derivative of d^2.... 2dd'?

Answer 32

or just 2d?

Answer 33

don't take the derivative of d^2 , name it as f(x)

Answer 34

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

Answer 35

ok. I see what u did there. but I don't see why.

Answer 36

minimizing distance and square of distance is same thing!
u get same answer for both, only that minimizing SQUARE of distance is a lot easier .....

Answer 37

ok. then. I got \[f'(x)=2(x-2)+2x(x ^{2}-\frac{ 1 }{ 2 })\]

Answer 38

2x or 4x ?

Answer 39

Oops. 4x (/.-)

Answer 40

yes, now equate that to 0 and find values of x

Answer 41

How do I find the x's?

Answer 42

\(f'(x)=2(x-2)+4x(x ^{2}-\frac{ 1 }{ 2 })=0\)
do algebraic simplification

Answer 43

1?

Answer 44

u get a cubic equation.

Answer 45

yeah, 1 is a solution...

Answer 46

So the answere is (1,1) :D

Answer 47

i agree.

Answer 48

yep. thats the point

Answer 49

i wonder how you got x=1 so fast

Answer 50

Thanks @hartnn ! without u, I woudlnt have an A in calculus xD

Answer 51

I just got f(1)

Answer 52

and that 1^2=1

Answer 53

ok.... welcome ^_^