Question

Find any critical numbers of the function. $$ f(x)=\frac{4 x}{x^{2}+1} $$

Short answer

The critical numbers of the function \(f(x) = \frac{4x}{x^2 + 1}\) are \(x = 1\) and \(x = -1\).

Step-by-step solution

Find the Derivative

The first step in finding the critical numbers is to find the derivative of the function. Based on the Quotient Rule, the derivative of the function \(f(x) = \frac{4x}{x^2 + 1}\) is:\(f'(x) = \frac{(x^2+1)(4)-(4x)(2x)}{(x^2 + 1)^2}\)

Simplify the Derivative

Next, simplify the derivative function by multiplying the terms and combining similar ones:\(f'(x) = \frac{4x^2+4-8x^2}{(x^2 + 1)^2} = \frac{-4x^2+4}{(x^2 + 1)^2}\)

Find Where Derivative Equals Zero

Next, the derivative should be set to zero and solve the equation to figure out where it equals zero:\(\frac{-4x^2+4}{(x^2 + 1)^2} = 0\)This applies when the numerator of the fraction is equal to zero: \(-4x^2+4 = 0\)

Solve the Equation

Solving:\(-4x^2 + 4 = 0\)Leads to:\(x = \sqrt{1} = ±1\)

Check Where Function Doesn't Exist

The function \(f(x) = \frac{4x}{x^2 + 1}\) exists for all values of x. Therefore, there are no x-values that make function undefined.