given f(x)=�sqrt(x). find the distance of the point P(2,5) from the line tangent to the functions at x=4.

Answer 1

y(x) = √x
y'(x) = 1/(2√x)

y'(4) = 1/(2√4) = 1/4

So, at (4,2) the tangent line is

(y-2)/(x-4) = 1/4
4y - 8 = x - 4
-x + 4y - 4 = 0

The distance from (m,n) to the line Ax+By+C=0 is

|mA+nB+C|/√(A62 + B^2)

The distance from (2,5) to the line is thus

|-1*2 + 4*5 - 4|/√(1+16) = |-2+20-4|/√17 = 14/√17