If the line 3x-y+2=0 is tangent in the first quadrant to the curve y=x^3+k, then k= ?

Answer 1

For the given function, dy/dx = 3x^2. The line is recast as y = 3x + 2, which has a slope of 3.
For the line to be a tangent, 3 = 3x^2, or x= 1 in the first quadrant. For the line and curve to be equal at x= 1, for the line, y= 5. Then, for the curve to have the coordiante (1,5), k=2.

Answer 2

If the line 3x - y + 2 = 0 is tangent in the first quadrant to the curve y = x³ + k, then k = ?

The equation of the line is:
3x - y + 2 = 0
y = 3x + 2

The derivative of the curve is:
dy/dx = 3x²

For the line to be tangent to the curve the derivative must equal the slope of the line, which is 3. And in the first quadrant that occurs at:

dy/dx = 3x² = 3
x² = 1
x = 1

Solve for x = 1 on the line.

y = 3x + 2 = 3*1 + 2 = 3 + 2 = 5

That means the curve y = x³ + k passes thru the point (1,5).

y = x³ + k
5 = 1³ + k
k = 4