What we are being asked here is to simply minimize distance. Also, note that we can write f(x)=√x as y=√x.
Now, what is this "distance?" How do we find it? Well, if you think back to Algebra I or Geometry, you'll remember that the distance between two points (x1,y1) and (x2,y2) is given by: √(y2−y1)2+(x2−x1)2. For example, the distance between the points (4,0) and (0,3) would be:
√(3−0)2+(4−0)2=√9+16=√25=5
Ok, so what is (x1,y1) and (x2,y2) in our example? (x1,y1) is simple - it's just the point given in the problem, (4,0). Because we don't know what x2 is, we'll just call it x for now. As for y2, we don't know that either; and since y=√x, we'll call it √x.
Our formula then becomes:
√(√x−0)2+(x−4)2=√(√x2)+x2−8x+16=√x+x2−8x+16=√x2−7x+16
We are being asked to minimize this distance, which we'll call s to make the following calculations easier. To minimize something, we have to take its derivative, so let's start there:
s=√x2−7x+16=(x2−7x+16)12
dsdx=(2x−7)⋅12(x2−7x+16)12→Using power rule and chain rule
dsdx=2x−72√x2−7x+16
Now we set this equal to 0 and solve for x:
0=2x−72√x2−7x+16
0=2x−7
x=72
This is known as the critical value, and it represents the x-value for which the function is minimized. All we need to do now is find the corresponding y-value, using the definition of y: y=√x. Substituing 72 for x:
y=√72
y≈1.87
And voila, the y-value. We can now say that the minimum distance between f(x)=√x and the point (4,0) (the place where these two are closest) occurs at (72,1.87). For a little extra fun, we can use the distance formula to see what the actual distance between the points is:
s=√(1.87−0)2+(72−4)2≈1.8 units
