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

Question

a: 1/2
b: 1/4
c: 0
d: -1/8
e: -1/2

Answer 1

if y = x^3 + k
dy/dx = 3x^2 , which of course is the slope of the tangent

but the slope of the given line is 3/4 , so ...

3x^2 = 3/4
x^2 = 1/4
x = ± 1/2 , but since we are restricted to the first quad,
x = 1/2
if x=1/2, 3(1/2) - 4y = 0
y = 3/8

The point of contact of the tangent is (1/2 , 3/8)
so in the cubic ...
3/8 = (1/8) + k
k = 3/8-1/8 = 1/4

Answer 2

Thank youu(: