A curve is defined by the equation y = 2x^3 - 9x^2 + 12x + 5. What is the value of y at it… — JAMB UTME

A curve is defined by the equation $y = 2x^3 - 9x^2 + 12x + 5$ . What is the value of $y$ at its maximum stationary point?

Answer: 10

The value of

y

at the maximum stationary point is

10

To locate the maximum stationary point, set the first derivative

\frac{dy}{dx} = 6x^2 - 18x + 12

to zero, obtaining stationary values

x = 1

and

x = 2

. Testing in the second derivative

\frac{d^2y}{dx^2} = 12x - 18

gives

-6

x = 1

, confirming a local maximum. Substituting

x = 1

into the original cubic equation gives

y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 10

Differentiate

y

with respect to

x

to find the gradient function.

\frac{dy}{dx} = 6x^2 - 18x + 12

Stationary points occur where the gradient of the curve is zero.

Set the first derivative to zero and solve for

x

x = 1 \text{ or } x = 2

Factoring

6(x - 1)(x - 2) = 0

yields the

x

-coordinates of the turning points.

Determine the nature of the stationary points using the second derivative test.

\frac{d^2y}{dx^2} = 12x - 18; \quad \text{at } x = 1, \frac{d^2y}{dx^2} = -6 < 0

A negative second derivative indicates a local maximum stationary point.

Substitute

x = 1

into the original function to determine

y

y = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 10

Evaluating the curve function at the maximum

x

-coordinate provides the corresponding maximum

y

-value.

Stationary Points, Maxima, and Minima

Question