# Find the positive critical point of the function f(x) = \frac{x}{x^2 + 3}.

## Question:

Find the positive critical point of the function {eq}f(x) = \frac{x}{x^2 + 3}. {/eq}

## Critical Points:

Critical points of differentiable functions at all points are reduced to a single group:

the points for which the derivative of the function is equal to zero.

Given the function, {eq}f(x) = \frac{x}{x^2 + 3} {/eq}, taking into account the domain is given by all the real numbers (the denominator is not equal form zero):

{eq}f(x) = \frac{x}{{{x^2} + 3}}\\ \\ {x^2} + 3 \ne 0\\ D\left( f \right) = R {/eq}

The critical point is given only by the numbers where the derivative is equal to zero, that is:

{eq}f'(x) = \frac{{1\left( {{x^2} + 3} \right) - x \cdot 2x}}{{{{\left( {{x^2} + 3} \right)}^2}}}\\ f'(x) = \frac{{{x^2} + 3 - 2{x^2}}}{{{{\left( {{x^2} + 3} \right)}^2}}}\\ f'(x) = \frac{{ - {x^2} + 3}}{{{{\left( {{x^2} + 3} \right)}^2}}} = 0\\ \\ - {x^2} + 3 = 0\\ {x^2} = 3\\ x = \pm \sqrt 3 {/eq}