Posted by Susana on Monday, October 3, 2011 at 12:57pm.
Find k such that the line is tangent to the graph of the function. Function: f(x)=kã(x)
Line: y=x+4

Calculus  Reiny, Monday, October 3, 2011 at 2:35pm
I will assume your equation is
f(x) = k√x = k x^(1/2)
then f'(x) = (k/2)x^(1/2)
but the slope of y = x+4 is 1
so (k/2)x^(1/2) = 1
k/(2√x) = 1
k = 2√x
then y = 2√x √x = 2x
sub back into y = x+4
2x = x+4
x = 4
so k= 2√4 = 4 
Calculus  Anonymous, Sunday, January 13, 2013 at 6:07am
y=2x²√ 2x