Question

Find the value of the constant k for which the line y = 2x + 1 is a tangent to the curve y = x^3 + kx + 3.

Answer 1

Cure is y = x³ + kx + 3.

Line is y = 2x + 1.

There must be one and more points that the line and the curve have in common. Means there are values of x for which the two functions are equal.

Algebrically x³ + kx + 3 = 2x + 1 ->(1)

Derivative of the curve is the slope of the tangent line y = 2x + 1.

y' = 3x² + k

2 = 3x² + k                                                 [Since slope = y' = 2]

k = 2 - 3x²   ->(2)

Substitute (2) in (1).

x³ + kx + 3 = 2x + 1

x³ + (2 - 3x²)x + 3 = 2x + 1

x³ + 2x - 3x³ + 3 = 2x + 1

-2x³ + 2x + 3 = 2x + 1

-2x³ + 2x + 3 - 2x - 1 = 0

-2x³ + 2 = 0

-2x³ = -2

x³ = -2/-2

x³ = 1

x = ± ∛1

x = ± 1

Substiute x values in equation (2).

k = 2 - 3(1)² = 2 - 3 = -1

k = 2 - 3(-1)² = 2 - 3 = -1

Therefore the value of k is -1.