Is #f(x)=(x-3)(x+5)(x+2)# increasing or decreasing at #x=-3#?

Question

Ezra Adams · Accepted Answer

The function is decreasing.

The answer lies if the first derivative. If it is positive at #x=-3#, the function is increasing, if it is negative, it's decreasing. It's easier to compute the derivative of the expanded form, and we have that #(x-3)(x+5)(x+2)=x^3+4 x^2-11 x-30# Thus, the initial derivative is #f'(x)=3x^2+8x-11# Evaluating #f'(-3)#, we have #3(-3)^2 +8(-3)-11 = 3*9 -24-11 = 27-24-11=-8# Thus, the function is getting smaller.

Isabella Ackerman · Answer

undefined To determine if the function $ f(x) = (x-3)(x+5)(x+2) $ is increasing or decreasing at $ x = -3 $, we need to examine the sign of the derivative of the function at that point.

First, find the derivative of the function $ f(x) $ using the product rule:
$$ f'(x) = (x+5)(x+2) + (x-3)(x+2) + (x-3)(x+5) $$

Now, plug in $ x = -3 $ into the derivative:
$$ f'(-3) = (-3+5)(-3+2) + (-3-3)(-3+2) + (-3-3)(-3+5) $$
$$ = (2)(-1) + (-6)(-1) + (-6)(2) $$
$$ = -2 + 6 - 12 $$
$$ = -8 $$

Since the derivative $ f'(-3) = -8 $ is negative, the function is decreasing at $ x = -3 $.

Axel Alvarez · Answer

undefined To determine if the function $ f(x) = (x-3)(x+5)(x+2) $ is increasing or decreasing at $ x = -3 $, we need to analyze the sign of the derivative of the function at that point.

The derivative of the function $ f(x) $ can be found using the product rule and is given by:

$$ f'(x) = (x+5)(x+2) + (x-3)(x+2) + (x-3)(x+5) $$

Evaluating the derivative at $ x = -3 $, we get:

$$ f'(-3) = (-3+5)(-3+2) + (-3-3)(-3+2) + (-3-3)(-3+5) $$
$$ f'(-3) = (2)(-1) + (-6)(-1) + (-6)(2) $$
$$ f'(-3) = -2 + 6 - 12 $$
$$ f'(-3) = -8 $$

Since the derivative $ f'(-3) $ is negative, the function $ f(x) $ is decreasing at $ x = -3 $.