A curve is defined by the equation y = 4x + \frac{9}{x} for x > 0. What is the y-value at … — JAMB UTME

A curve is defined by the equation $y = 4x + \frac{9}{x}$ for $x > 0$ . What is the $y$ -value at the minimum stationary point of the curve?

Answer: 12

The

y

-value at the minimum stationary point is 12.

To find the minimum value of

y = 4x + \frac{9}{x}

for

x > 0

, set the first derivative

\frac{dy}{dx} = 4 - \frac{9}{x^2}

equal to

0

, yielding

x = 1.5

. The second derivative

\frac{d^2y}{dx^2} = \frac{18}{x^3}

is positive at

x = 1.5

, confirming a minimum stationary point. Evaluating the original equation at

x = 1.5

gives

y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12

Differentiate the given function

y = 4x + 9x^{-1}

with respect to

x

\frac{dy}{dx} = 4 - 9x^{-2} = 4 - \frac{9}{x^2}

Stationary points occur where the gradient (first derivative) equals zero.

Set the first derivative to zero and solve for

x

given the domain constraint

x > 0

4 - \frac{9}{x^2} = 0 \implies 4x^2 = 9 \implies x^2 = \frac{9}{4} \implies x = \frac{3}{2} = 1.5

Solving

\frac{dy}{dx} = 0

yields the

x

-coordinate of the turning point.

Evaluate the second derivative to confirm the turning point is a local minimum.

\frac{d^2y}{dx^2} = \frac{18}{x^3}. \text{ At } x = 1.5, \frac{d^2y}{dx^2} = \frac{18}{3.375} = 5.333 > 0

A positive second derivative indicates that the stationary point is a local minimum.

Substitute

x = 1.5

back into the original equation

y = 4x + \frac{9}{x}

to find the corresponding

y

-value.

y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12

The question asks for the

y

-value of the curve at the minimum stationary point.

Stationary Points, Maxima, and Minima

Estimated Time:2m 0s

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