How do you find the critical points of the function #f(x) = x / (x^2 + 4)#?

Question

Cameron Alexander · Accepted Answer

Find those points at which the derivative of #f# is equal to 0

The critical points of a function are those points where its first derivative is 0, i.e. those points where the function reaches a maximum, a minimum, or a point of inflection. In this case, #f(x)=x/(x^2+4)#, so #f'(x)=(4-x^2)/(x^2+4)^2# by the quotient rule (and a little combining of terms). This equals 0 either when the denominator equals #oo# (which doesn't happen here for non-infinite #x#) or when the numerator equals 0. So we want #4-x^2=0#, which tells us the two critical points of the function: #x=+-2#, which equate to #f(x)=+-1/4#.

Aurora Alvarado · Answer

undefined To find the critical points of the function $ f(x) = \frac{x}{x^2 + 4} $, follow these steps:

1. Find the derivative of the function $ f(x) $ with respect to $ x $, denoted as $ f'(x) $.
2. Set $ f'(x) $ equal to zero and solve for $ x $.
3. The solutions obtained in step 2 are the critical points of the function.

Let's go through these steps:

1. The derivative $ f'(x) $ can be found using the quotient rule:

$$ f'(x) = \frac{(x^2 + 4) \cdot 1 - x \cdot (2x)}{(x^2 + 4)^2} $$

Simplifying this expression yields:

$$ f'(x) = \frac{x^2 + 4 - 2x^2}{(x^2 + 4)^2} $$
$$ f'(x) = \frac{-x^2 + 4}{(x^2 + 4)^2} $$

2. Set $ f'(x) $ equal to zero and solve for $ x $:

$$ \frac{-x^2 + 4}{(x^2 + 4)^2} = 0 $$

Since the numerator can't be zero (it's always $ -x^2 + 4 $), the critical points occur when the denominator is zero:

$$ x^2 + 4 = 0 $$

This equation has no real solutions, so there are no critical points for the function $ f(x) = \frac{x}{x^2 + 4} $.