A curve is given by the equation y = \frac{x}{x^2 + 4}. What is the maximum value of y on … — JAMB UTME

A curve is given by the equation $y = \frac{x}{x^2 + 4}$ . What is the maximum value of $y$ on this curve?

The maximum value of

y

on the curve is

\frac{1}{4}

To find the maximum value of

y = \frac{x}{x^2 + 4}

, differentiate

y

using the quotient rule to obtain

\frac{dy}{dx} = \frac{4 - x^2}{(x^2 + 4)^2}

. Setting the numerator to zero gives

x = \pm 2

. Since

\frac{dy}{dx}

is positive for

x < 2

and negative for

x > 2

x = 2

corresponds to a local maximum. Substituting

x = 2

into

y = \frac{x}{x^2 + 4}

yields

y = \frac{2}{8} = \frac{1}{4}

Differentiate

y = \frac{x}{x^2 + 4}

using the quotient rule

\frac{dy}{dx} = \frac{(x^2 + 4)(1) - x(2x)}{(x^2 + 4)^2} = \frac{4 - x^2}{(x^2 + 4)^2}

Stationary points occur where the first derivative

\frac{dy}{dx}

equals zero.

Find the stationary points by setting

\frac{dy}{dx} = 0

4 - x^2 = 0 \implies x = 2 \text{ or } x = -2

A fraction equals zero when its numerator is zero.

Determine the nature of the stationary point at

x = 2

\text{For } x < 2, \frac{dy}{dx} > 0; \text{ for } x > 2, \frac{dy}{dx} < 0 \implies x = 2 \text{ is a maximum point}

The derivative changes sign from positive to negative across a maximum point.

Evaluate

y

x = 2

y = \frac{2}{2^2 + 4} = \frac{2}{8} = \frac{1}{4}

Substituting the

x

-coordinate into the curve equation yields the maximum value of

y

Stationary Points, Maxima, and Minima

Question