## MarcLeclair 2 years ago Find the point on the graph of √(x+1) that lies closest to point ( 4,0).

1. galanh

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

2. MarcLeclair

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

3. MarcLeclair

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

4. galanh

the closed point is (3.5,2.12)

5. MarcLeclair

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

6. MarcLeclair

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

7. galanh

I used this graph

8. MarcLeclair

9. MarcLeclair

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

10. galanh

Thanks for the solution with derivative.

11. MarcLeclair

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

12. galanh

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

13. MarcLeclair

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

14. galanh

My pleasure!