A curve is defined by the equation y = x^3 - 3x^2 + k, where k is a constant. If the local… — JAMB UTME

A curve is defined by the equation $y = x^3 - 3x^2 + k$ , where $k$ is a constant. If the local minimum value of $y$ on the curve is $2$ , what is the value of $k$ ?

Answer: 6

Answer

The value of the constant

k

6

To find the constant

k

, differentiate the curve equation to obtain

\frac{dy}{dx} = 3x^2 - 6x

. Setting

\frac{dy}{dx} = 0

gives stationary points at

x = 0

and

x = 2

. Calculating the second derivative

\frac{d^2y}{dx^2} = 6x - 6

shows

\frac{d^2y}{dx^2} = 6 > 0

x = 2

, confirming that the local minimum occurs at

x = 2

. Substituting

x = 2

and the minimum value

y = 2

into

y = x^3 - 3x^2 + k

yields

2 = 8 - 12 + k

, which simplifies to

k = 6

Step-by-Step Solution

Differentiate

y = x^3 - 3x^2 + k

with respect to

x

\frac{dy}{dx} = 3x^2 - 6x

Stationary points occur where the first derivative equals zero.

Solve

\frac{dy}{dx} = 0

for

x

x = 0

x = 2

These

x

-values locate the turning points on the curve.

Evaluate the second derivative

\frac{d^2y}{dx^2} = 6x - 6

at each stationary point.

x = 0

\frac{d^2y}{dx^2} = -6

(local maximum); at

x = 2

\frac{d^2y}{dx^2} = 6

(local minimum).

A positive second derivative indicates a local minimum point.

Substitute

x = 2

and

y = 2

into the curve equation

y = x^3 - 3x^2 + k

2 = (2)^3 - 3(2)^2 + k \implies 2 = -4 + k \implies k = 6

The local minimum value of

y

is attained at

x = 2

Key Concept

Stationary Points, Maxima, and Minima

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