What are the critical numbers for #f(x)= x/(x+1)#?

Question

Aurora Alvarado · Accepted Answer

This function has no critical numbers.

A critical number for #f# is a number in the domain of #f# at which the derivative is either #0# or fails to exist. For #f(x) = x/(x+1)#, the derivative is: #f'(x) = 1/(x+1)^2# #f'(x)# is never #0# and #f'(x)# is undefined at #x=-1#, but #-1# is not in the domain of #f#, So, #f# has no critical numbers.

Christopher Allers · Answer

undefined To find the critical numbers of $ f(x) = \frac{x}{x+1} $, you need to find where the derivative of the function is equal to zero or undefined.

1. Find the derivative of $ f(x) $ using the quotient rule.
$$ f'(x) = \frac{(x+1)(1) - (x)(1)}{(x+1)^2} $$
$$ f'(x) = \frac{x+1 - x}{(x+1)^2} $$
$$ f'(x) = \frac{1}{(x+1)^2} $$

2. Set the derivative equal to zero and solve for $ x $.
$$ \frac{1}{(x+1)^2} = 0 $$
Since a fraction can only be zero if its numerator is zero, the denominator can't be zero.
$$ 1 = 0 $$
There are no solutions for $ x $ in this case.

3. Determine where the derivative is undefined. The derivative is undefined when the denominator is zero.
$$ (x+1)^2 = 0 $$
$$ x+1 = 0 $$
$$ x = -1 $$

So, the critical number for $ f(x) = \frac{x}{x+1} $ is $ x = -1 $.