Find the point on the graph of the function that is closest to the given point.
Function Point
f(x) = x^2 (16, 1/2)
(x, y) = ( ) ?
Let D be the distance from a point on the function line from (16, 1/2)
D^2 = (x-16)^2 + (y - 1/2)^2
= (x-16)^2 + (x^2 -1/2)^2
When this is a minimum, dD^2/dx = 0
2(x-16)+ 2(x^2 -1/2)*2x = 0
2x - 32 + 4x^3 -2x = 0
4x^3 = 32
x = +/- 2
y = 4
Thanks :)
To find the point on the graph of the function that is closest to the given point, we can use the formula for the distance between two points in a coordinate plane.
The given point is (16, 1/2). Let's assume the point on the graph of the function is (x, y). We need to find the value of x and y that minimizes the distance between the two points.
The distance formula is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Substituting the coordinates of the given point and the unknown point into the distance formula, we get:
d = sqrt((x - 16)^2 + (y - 1/2)^2)
To minimize the distance, we need to minimize the expression inside the square root. So, we can minimize the squared expression:
(x - 16)^2 + (y - 1/2)^2
Now, we differentiate the expression with respect to x and y separately and set the derivatives equal to zero to find the critical points.
Differentiating with respect to x:
d/dx ((x - 16)^2 + (y - 1/2)^2) = 2(x - 16)
Setting the derivative equal to zero:
2(x - 16) = 0
x - 16 = 0
x = 16
Differentiating with respect to y:
d/dy ((x - 16)^2 + (y - 1/2)^2) = 2(y - 1/2)
Setting the derivative equal to zero:
2(y - 1/2) = 0
y - 1/2 = 0
y = 1/2
We have found a critical point at (x, y) = (16, 1/2).
Therefore, the point on the graph of the function that is closest to the given point (16, 1/2) is (16, 1/2).