Find k such that the line is tangent to the graph of the function. Function: f(x)=k�ã(x)

Question

Line: y=x+4

Answer 1

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

Answer 2

y=2x²√ 2-x