Is #f(x)=(x+7)(x-2)(x-1)# increasing or decreasing at #x=-1#?

Question

William Abdalla · Accepted Answer

It is increasing if the derivative at #x=-1# is positive and decreasing if the derivative at #x=-1# is negative.

In order to answer this question you need to find #f'(x)# There are a few different ways to go about this, but this is the one I went with so that I could quickly apply the product rule. I would advise using algebra to simplify the function before taking the derivative. #f(x)=(x+7)(x^2-3x+2)# #f'(x)=1(x^2-3x+2)+(x+7)(2x-3)# #f'(-1)=1*((-1)^2-3*(-1)+2)+(-1+7)(2*(-1)-3)# #f'(-1)=7+(-30)=-23# #f'(-1)<0# therefore the function is decreasing at #x=-1#

Isabella Ackerman · Answer

undefined To determine if $ f(x) = (x + 7)(x - 2)(x - 1) $ is increasing or decreasing at $ x = -1 $, we can check the sign of the derivative at that point. We find the derivative of $ f(x) $ and then evaluate it at $ x = -1 $.

The derivative of $ f(x) $ is:

$$ f'(x) = (x - 2)(x - 1) + (x + 7)(x - 1) + (x + 7)(x - 2) $$

To evaluate the sign of $ f'(x) $ at $ x = -1 $, we substitute $ x = -1 $ into the derivative expression.

$$ f'(-1) = (-1 - 2)(-1 - 1) + (-1 + 7)(-1 - 1) + (-1 + 7)(-1 - 2) $$
$$ f'(-1) = (-3)(-2) + (6)(-2) + (6)(-3) $$
$$ f'(-1) = 6 - 12 - 18 $$
$$ f'(-1) = -24 $$

Since $ f'(-1) = -24 $ is negative, the function $ f(x) $ is decreasing at $ x = -1 $.