Question

Find the point on the graph of √(x+1) that lies closest to point ( 4,0).

Answer 1

The distance between a point on the curve and the point (0,4) is:
\[d= \sqrt{(x-4)**2+x+1}=\sqrt{x**2 -7x+17}\]
\[d=\sqrt{(x-3.5)**2 +4.75}\]
d is minimum when x=3.5
and y=sqrt(3.5+1)=sqrt(4.5)=2.12

Answer 2

Is that the formula for the distance of a curve to a point, the one they teach in grade 10 or something?

Answer 3

we dont want the distance, we want the point closest, and the answer is wrong :/

Answer 4

the closed point is (3.5,2.12)

Answer 5

wait... is that a basic equation? it's for my calculus class and I didn't see you use any derivative or anything

Answer 6

@galanh its odd my solution sheet has a different way, I'm just trying to understand the method sorry :/

Answer 7

I used this graph

Answer 8

thats the solution using derivative, but I don't understand why we would set the first derivative to 0 to find the critical points, just seems odd to me

Answer 9

Thanks for the solution with derivative.

Answer 10

do you understand that solution? I don't understand it that was why i asked the question in the first place

Answer 11

I am not good with derivatives!
derivative of a function equal to zero, gives you the minimum of the function!

Answer 12

Errr thanks for the help :) i'll try to figure out the rest on my own haha

Answer 13

My pleasure!