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

Question

Ezra Adams · Accepted Answer

Increasing.

Find the first derivative at the given point. If the value is positive, the function is increasing. If the value is negative the function is decreasing. Finding #f'(x)# will require the quotient rule: #f'(x)=((x+3)d/dx[x^3+2x^2-x+2]-(x^3+2x^2-x-2)d/dx[x+3])/(x+3)^2# #=((x+3)(3x^2+4x-1)-(x^3+2x^2-x-2))/(x+3)^2# #=(3x^3+13x^2+11x-3-(x^3+2x^2-x-2))/(x+3)^2# #=(2x^3+11x^2+12x-1)/(x+3)^2# Now, calculate the derivative at the point. #f'(-2)=(2(-8)+11(4)+12(-2)-1)/(-2+3)^2# #=(-16+44-24-1)/1# #=3# Since #3>0#, the function is increasing when #x=-2#.

William Abdalla · Answer

undefined To determine whether the function $ f(x) = \frac{x^3 + 2x^2 - x - 2}{x + 3} $ is increasing or decreasing at $ x = -2 $, we can use the first derivative test.

First, we find the derivative of $ f(x) $:
$$ f'(x) = \frac{d}{dx}\left(\frac{x^3 + 2x^2 - x - 2}{x + 3}\right) $$

Using the quotient rule, we have:
$$ f'(x) = \frac{(x + 3)(3x^2 + 4x - 1) - (x^3 + 2x^2 - x - 2)(1)}{(x + 3)^2} $$

Simplify the expression:
$$ f'(x) = \frac{3x^3 + 4x^2 - x + 9x^2 + 12x - 3 - x^3 - 2x^2 + x + 2}{(x + 3)^2} $$
$$ f'(x) = \frac{2x^3 + 11x^2 + 12x - 1}{(x + 3)^2} $$

Now, we find the value of $ f'(x) $ at $ x = -2 $:
$$ f'(-2) = \frac{2(-2)^3 + 11(-2)^2 + 12(-2) - 1}{(-2 + 3)^2} $$
$$ f'(-2) = \frac{-16 + 44 - 24 - 1}{1} $$
$$ f'(-2) = 3 $$

Since $ f'(-2) > 0 $, the function $ f(x) $ is increasing at $ x = -2 $.