If the line 3x-4y = 0 is tangent in the first quadrant to the curve y = x^3 + k, then k is

Answer 1

4 y = 3 x + 0
y = (3/4) x
slope = m = 3/4

y = x^3 + k
dy/dx = slope = 3 x^2
so
3 x^2 = 3/4
x^2 = 1/4
x = 1/2 (-1/2 is not in first quadrant)
then
y = (1/2)^3 + k
y = 1/8 + k and y = 3x/4
so
3 x/4 = 1/8 + k
but x = 1/2
3/8 = 1/8 + k
k = 2/8 = 1/4