Is #f(x)=(1-x)/(x+2)^3# increasing or decreasing at #x=-1#?

Question

Charles Anctil · Accepted Answer

The function is increasing at #x = -1#

Differentiate the function to get the gradient. If the gradient is positive the function is increasing, and if it is negative the function is decreasing. #f(x) = (1-x)/(x+2)^3 # #f'(x) = (-x(x+2)^3 -(1-x)(3(x+2)^2))/(x+2)^6# #=(-x(x+2)^3 - 3(x+2)^2 + 3x(x+2)^2)/(x+2)^6# #=((x+2)^2(-x(x+2)-3 +3x))/(x+2)^6# #=(-x^2-2x-3+3x)/(x+2)^4# #=(-x^2+x+3)/(x+2)^4# At #x = -1# this becomes #(-(-1)^2 +(-1) +3)/(-1+2)^4 = (-1-1+3)/(1)^4# #=1#

Since this is a positive, the function is growing.

Ezra Adams · Answer

undefined To determine if the function $ f(x) = \frac{1-x}{(x+2)^3} $ is increasing or decreasing at $ x = -1 $, we need to analyze the sign of its derivative at that point. We can find the derivative of $ f(x) $ with respect to $ x $ and then evaluate it at $ x = -1 $.

The derivative of $ f(x) $ is:

$$ f'(x) = \frac{d}{dx} \left( \frac{1-x}{(x+2)^3} \right) $$

To find $ f'(x) $, we can use the quotient rule:

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

Now, we substitute $ x = -1 $ into $ f'(x) $:

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

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

$$ f'(-1) = \frac{-1 - 6}{1} $$

$$ f'(-1) = -7 $$

Since the derivative $ f'(-1) = -7 $ is negative, the function $ f(x) = \frac{1-x}{(x+2)^3} $ is decreasing at $ x = -1 $.