Is #f(x)=(x+3)(x-6)(x/3-1)# increasing or decreasing at #x=-2#?

Question

Paisley Johnson · Accepted Answer

#"increasing at "x=-2# #"To determine if "f(x)" is increasing/decreasing at"# #"x = -2 differentiate and evaluate at x = - 2"# #• " if "f'(-2)>0" then "f(x)" is increasing at x = - 2"# #• " if "f'(-2)<0" then "f(x)" is decreasing at x = - 2"# #"expanding the factors gives"# #f(x)=1/3x^3-2x^2-3x+18# #rArrf'(x)=x^2-4x-3# #rArrf'(-2)=4+8-3=9>0# #"Since "f'(-2)>0" then "f(x)" is increasing at x = - 2"# graph{(x+3)(x-6)(1/3x-1) [-20, 20, -10, 10]}

Samantha Adkison · Answer

undefined To determine whether the function $ f(x) = (x + 3)(x - 6)\left(\frac{x}{3} - 1\right) $ is increasing or decreasing at $ x = -2 $, we can analyze the sign of the derivative of the function at that point. If the derivative is positive, the function is increasing; if negative, it's decreasing.

We'll first find the derivative of the function and then evaluate it at $ x = -2 $.

$$ f(x) = (x + 3)(x - 6)\left(\frac{x}{3} - 1\right) $$

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

Now, evaluate $ f'(-2) $ to determine whether the function is increasing or decreasing at $ x = -2 $.

Axel Alvarez · Answer

undefined To determine if the function $ f(x) = (x + 3)(x - 6)\left(\frac{x}{3} - 1\right) $ is increasing or decreasing at $ x = -2 $, we need to evaluate the first derivative of the function and determine its sign at $ x = -2 $.

First, let's find the derivative $ f'(x) $ of $ f(x) $:

Using the product rule for differentiation:

$$ f'(x) = \left(1\right)(x - 6)\left(\frac{x}{3} - 1\right) + (x + 3)\left(1\right)\left(\frac{x}{3} - 1\right) + (x + 3)(x - 6)\left(\frac{1}{3}\right) $$

Simplifying:

$$ f'(x) = (x - 6)\left(\frac{x}{3} - 1\right) + (x + 3)\left(\frac{x}{3} - 1\right) + \frac{1}{3}(x + 3)(x - 6) $$

Now, evaluate $ f'(x) $ at $ x = -2 $:

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

$$ f'(-2) = (-8)\left(\frac{-2}{3} - 1\right) + (1)\left(\frac{-2}{3} - 1\right) + \frac{1}{3}(1)(-8) $$

$$ f'(-2) = (-8)\left(\frac{-2 - 3}{3}\right) + (1)\left(\frac{-2 - 3}{3}\right) - \frac{8}{3} $$

$$ f'(-2) = (-8)\left(\frac{-5}{3}\right) + (1)\left(\frac{-5}{3}\right) - \frac{8}{3} $$

$$ f'(-2) = \frac{40}{3} - \frac{5}{3} - \frac{8}{3} $$

$$ f'(-2) = \frac{27}{3} $$

$$ f'(-2) = 9 $$

Since $ f'(-2) = 9 $ is positive, the function $ f(x) $ is increasing at $ x = -2 $.