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Question 528114: which point of the graph f(x)=x^2 are the closest to the point (2, 0.5)?
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! which point of the graph f(x)=x^2 are the closest to the point (2, 0.5)?
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One way to solve is to find the slope of the tangent line.
The line passing through the point (2, 0.5) and having the minimum distance to
f(x) will perpendicular to the tangent line (normal line).
Let the point on the graph = (x0,y0)
The slope of the tangent line = df(x)/dx = 2x = 2x0 at the point x0
The normal line is perpendicular to the tangent line, so its slope is the neg. reciprocal:
Slope(normal) = -1/(2x0)
But the slope of a line going through the points (x0,y0) and (2,0.5) is:
(y0-0.5)/(x0-2)
We can equate the two expressions for the slope:
-1/(2x0) = (y0-0.5)/(x0-2)
But y0 = x0^2 from the formula for the graph:
Substitute this value and cross-multiply to solve for x0:
x0^3 = 1 -> x0 = 1
And since y0 = x0^2 -> y0 = 1
So the point is (1,1)
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