What is the value of y at the local minimum stationary point of the curve y = x^3 - 3x^2 -… — JAMB UTME

What is the value of $y$ at the local minimum stationary point of the curve $y = x^3 - 3x^2 - 9x + 15$ for $x > 0$ ?

Answer: -12

The value of

y

at the local minimum stationary point is

-12

To find the local minimum point of

y = x^3 - 3x^2 - 9x + 15

, set the derivative

\frac{dy}{dx} = 3x^2 - 6x - 9

equal to

0

. Solving

3(x - 3)(x + 1) = 0

yields

x = 3

for

x > 0

. The second derivative

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

equals

12 > 0

x = 3

, confirming a local minimum. Substituting

x = 3

back into the original curve equation yields

y = (3)^3 - 3(3)^2 - 9(3) + 15 = -12

Differentiate

y = x^3 - 3x^2 - 9x + 15

with respect to

x

\frac{dy}{dx} = 3x^2 - 6x - 9

Stationary points occur where the slope (first derivative) is zero.

Set

\frac{dy}{dx} = 0

and solve for

x

x = 3 \text{ or } x = -1

Factoring

3(x^2 - 2x - 3) = 0

gives

(x - 3)(x + 1) = 0

Apply the second derivative test at

x = 3

(since

x > 0

)

\frac{d^2y}{dx^2} = 6(3) - 6 = 12 > 0

A positive second derivative confirms that

x = 3

is a local minimum.

Evaluate

y

x = 3

in the original curve equation

y = (3)^3 - 3(3)^2 - 9(3) + 15 = -12

Substituting

x = 3

into

y(x)

gives the

y

-coordinate of the minimum point.

Stationary Points and Local Minima of Polynomial Curves

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