How do you find the critical numbers for #(2-x)/(x+2)^3# to determine the maximum and minimum?

Question

William Abdalla · Accepted Answer

At #x = -2#, we have a global maximum of #infty# and a global minimum of #-infty#. A local minimum occurs at #x = 4#.

Critical numbers are the #x# values for which #f'(x) = 0#. These critical points may or may not be maximums/minimums. We first need to find the derivative. We do so using the quotient rule. #f(x) = (2-x)/(x+2)^3# #f'(x) = ( ((x+2)^3)(-1) - (2-x)(3)(x+2)^2)/(x+2)^6# #f'(x) = ( -(x+2)^3 - 3(2-x)(x+2)^2 ) / (x+2)^6# Set this equal to zero and find which values of #x# satisfy the equation. #(-(x+2)^3 - 3(2-x)(x+2)^2)/(x+2)^6 = 0# #-(x+2)^3 - 3(2-x)(x+2)^2 = 0# #-(x+2)^3 = 3(2-x)(x+2)^2# #-(x+2) = 3(2-x)# #-x - 2 = 6 - 3x# #2x = 8# #x = 4# Note that #x = -2# is also a critical number, since the denominator of #f'(x)# will be #0# when #x = -2#. Investigating this critical number, we see that as #x -> -2# from the right, the denominator of #f(x)# gets very small and both the numerator and the denominator remain positive. This implies that the limit of #f(x)# as #x -> -2# from the right explodes to positive infinity. Similarly, as #x -> -2# from the left, #f(x)# explodes to negative infinity. Thus, the global maximum and minimum of the function are #infty# and #-infty#, respectively. Now let's investigate the critical number #x = 4#. We see that: #f'(0) = (-8 - 3(2)(8))/(64) = -56/(64)#. Also, #f'(5) = (-7^3 - 3(-3)(7^2))/(7^6) = (-343 + 441)/(7^6) = 98/7^6#. This tells us that before #x = 4#, the function is decreasing and after #x = 4#, the function is increasing. Thus, at #x = 4#, the function must have a local minimum.

Ezekiel Alexander · Answer

I have solved this way. Please, see the answer below:

Isaiah Allen · Answer

undefined To find the critical numbers of the function $ f(x) = \frac{2-x}{(x+2)^3} $, we first need to find the derivative of $ f(x) $ with respect to $ x $, then solve for $ x $ when the derivative is equal to zero or undefined.

The derivative of $ f(x) $ can be found using the quotient rule:

$$ f'(x) = \frac{d}{dx}\left(\frac{2-x}{(x+2)^3}\right) $$

$$ f'(x) = \frac{(x+2)^3(0 - 1) - (2-x)(3(x+2)^2)}{(x+2)^6} $$

$$ f'(x) = \frac{- (x+2)^3 + 3(2-x)(x+2)^2}{(x+2)^6} $$

$$ f'(x) = \frac{- (x+2)^3 + 3(2x+4)(x+2)^2}{(x+2)^6} $$

$$ f'(x) = \frac{- (x+2)^3 + 3(x+2)(2x+4)(x+2)}{(x+2)^6} $$

$$ f'(x) = \frac{- (x+2)^3 + 3(x+2)^2(2x+4)}{(x+2)^5} $$

To find the critical numbers, we set $ f'(x) $ equal to zero and solve for $ x $:

$$ - (x+2)^3 + 3(x+2)^2(2x+4) = 0 $$

$$ - (x+2)^3 + 3(2x+4)(x+2)^2 = 0 $$

After solving this equation, we can determine the values of $ x $ that correspond to the critical numbers of the function. Once we have these critical numbers, we can evaluate $ f(x) $ at these points as well as at the endpoints of any relevant intervals to determine the maximum and minimum values of the function.