Question

find the point on the graph of f(x)=x^2 that is closest to (4,6)

Answer 1

Let we are looking for the point (x,y) on f(x)=x^2 that is closest to (4,6)
=>y = x^2
The slope of the tangent at that point is
=>y' = 2x
The slope of the line between (x,y) and (4,6) = -1/y'
=>-1/2x = (y – 6)/(x-4)
=>-(x-4)=2xy-12x
=>2x^3-11x-4=0
=>x =2.5094
=>x=-2.1363
and x=-0.37308
So, (2.5094,6.29) is the point on f(x) closest to (4,6). To double check it graphically, it appears correct, as the smallest circle centered at (4,6) that touches f(x) seems to be at x=2.5094


Know more Sinusoidal Graph http://goo.gl/MOJm9U