Question

I need help with one last question, using distance. Find the point on the graph of f(x)=x^2 that is closest to the point (3,1/2)

Answer 1

use distance formula ..to write distance between (x,x^2) and (3, 0.5)

then find minimum by taking derivative

Answer 2

got the basic set up -> sqrt (x-2^2+(y-1/2)^2

Answer 3

oh

Answer 4

one sec, so it would be sqrt (x-3)^2+(y-1/2)^2

Answer 5

f'(x)=2x

Answer 6

replace y with x^2 in distance formula

Answer 7

okay, done

Answer 8

so that would be sqrt(x-3)^2+(x^2-1/2)^2

Answer 9

the set the derivative of the Distance formula to 0
... to find minimum distance

Answer 10

what is the y' of the distance formula. I'm confused

Answer 11

note: to make it easier.. you can get rid of the sqrt ... because the sqrt is at minimum when what's inside the sqrt is at minimum

Answer 12

good point
work with the square of the distance, not the distance

Answer 13

oh

Answer 14

but what would the derivative be?

Answer 15

would it be 2(x-3)+2(x^2-1/2) ?

Answer 16

\(D^2=(x-3)^2+(x^2-\frac{1}{2})^2\) multiply this mess out, combine like terms, take the derivative which should be easy enough because this is a polynomial

Answer 17

no

Answer 18

okay

Answer 19

gotta do the algebra first

Answer 20

gtg... my computer dying

Answer 21

i mean not really, but you are going to need to find the zeros of your derivative, so you will have to multiply out in any case

Answer 22

right now i hvae x^2+9+x^4+1/4

Answer 23

have*

Answer 24

okay, thank you so far for you help(:

Answer 25

oh no that is not it

Answer 26

i mean that is not what you get when you multiply out
you get this
http://www.wolframalpha.com/input/?i=%28x-3%29^2%2B%28x^2-1%2F2%29^2

Answer 27

oh okay

Answer 28

How would I simplify that?

Answer 29

oh , the y' is right there. okay. one sec

Answer 30

4x^3-6

Answer 31

(:

Answer 32

so would i set that to zero and solve?

Answer 33

yeah wolfram gave away the store on this one
you have the expression multiplied out and the derivative all in one

Answer 34

I see.

Answer 35

you even have the minimum there for you
\[4x^3-6=0\]
\[x^3=\frac{3}{2}\]
\[x=\sqrt[3]{\frac{3}{2}}\]

Answer 36

So that is the point closest to the point then. I'm looking at the process lol

Answer 37

thanks a bunch (:

Answer 38

http://www.wolframalpha.com/input/?i=expand+%28x-3%29%5E2%2B%28x%5E2-1%2F2%29%5E2