Question

Question #dbd28

Answer 1

Define the distance between the graph and the point as a function and find the minimum.

The point is `(3.5,1.871)`

Explanation:

To know how close they are, you need to know the distance. The Euclidean distance is:

`sqrt(Δx^2+Δy^2)`

where Δx and Δy are the differences between the 2 points. In order to be the nearest point, that point has to have the minimum distance. Therefore, we set:

`f(x)=sqrt((x-4)^2+(x^(1/2)-0)^2)`

`f(x)=sqrt(x^2-8x+16+(x^(1/2))^2)`

`f(x)=sqrt(x^2-8x+16+x^(1/2*2))`

`f(x)=sqrt(x^2-8x+16+x)`

`f(x)=sqrt(x^2-7x+16)`

We now need to find the minimum of this function:

`f'(x)=1/(2*sqrt(x^2-7x+16))*(x^2-7x+16)'`

`f'(x)=(2x-7)/(2*sqrt(x^2-7x+16))`

The denominator is always positive as a square root function. The numerator is positive when:

`2x-7>0`

`x>7/2`

`x>3.5`

So the function is positive when `x>3.5`. Similarly, it can be proved that it is negative when `x<3.5` Therefore, there function `f(x)` has a minimum at `x=3.5`, which means the distance is the least at `x=3.5` The y coordinate of `y=x^(1/2)` is:

`y=3.5^(1/2)=sqrt(3.5)=1.871`

Finally, the point where the least distance from (4,0) is observed is:

`(3.5,1.871)`