What is the maximum value of the curve y = \frac{x}{x^2 + 4}? — JAMB UTME

What is the maximum value of the curve $y = \frac{x}{x^2 + 4}$ ?

The maximum value of the curve is

\frac{1}{4}

Differentiating

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

using the quotient rule gives

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

. Setting

\frac{dy}{dx} = 0

yields critical points at

x = \pm 2

. Substituting

x = 2

into the original function gives

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

, which is the maximum value of the function.

Differentiate

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

with respect to

x

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 equals zero.

Set

\frac{dy}{dx} = 0

to solve for the stationary points.

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

A rational expression equals zero when its numerator is zero.

Evaluate

y

at each critical point to determine the function values.

For

x = 2

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

. For

x = -2

y = \frac{-2}{(-2)^2 + 4} = -\frac{1}{4}

The question asks for the maximum value of

y

on the curve.

Compare the stationary values to select the maximum.

The maximum value is

\frac{1}{4}

x = 2

Since

\frac{1}{4} > -\frac{1}{4}

x = 2

corresponds to the maximum point.

Stationary Points and Maxima/Minima of Rational Functions

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