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.

Answer and Explanation:

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}


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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