Is #f(x)=(-12x^2-22x-2)/(x-4)# increasing or decreasing at #x=1#?

Question

Emilia Aguilar · Accepted Answer

The function is increasing at #x=1# If #(dy)/(dx)# is positive then increasing, otherwise decreasing. Given:#" "color(brown)( y=(-12x^2-22x-2)/(x-4))# Using #y=u/v -> (dy)/(dx)= (v(du)/(dx)-u(dv)/(dx))/(v^2)# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #f'(x)=((x-4)(-24x-22)-(-12x^2-22x-2)(1))/((x-4)^2)# At #x=1# we have: #f'(x)=((1-4)(-24-22)-(-12-22-2)(1))/((1-4)^2)# #f'(x)= ("positive " +" possitive")/("possitive")-> "positive solution"# Thus as#f'(x)" "#is positive at #x=1# the function is increasing!

Axel Alvarez · Answer

undefined To determine whether $ f(x) = \frac{-12x^2 - 22x - 2}{x - 4} $ is increasing or decreasing at $ x = 1 $, we can use the first derivative test.

1. Calculate the first derivative of $ f(x) $, denoted as $ f'(x) $.
2. Substitute $ x = 1 $ into $ f'(x) $.
3. If $ f'(1) > 0 $, then $ f(x) $ is increasing at $ x = 1 $. If $ f'(1) < 0 $, then $ f(x) $ is decreasing at $ x = 1 $.

Let's find $ f'(x) $ first:

$$ f(x) = \frac{-12x^2 - 22x - 2}{x - 4} $$

Using the quotient rule:

$$ f'(x) = \frac{(x - 4)(-24x - 22) - (-12x^2 - 22x - 2)(1)}{(x - 4)^2} $$

Simplify $ f'(x) $:

$$ f'(x) = \frac{-24x^2 + 88x - 88 + 12x^2 + 22x + 2}{(x - 4)^2} $$
$$ f'(x) = \frac{-12x^2 + 110x - 86}{(x - 4)^2} $$

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

$$ f'(1) = \frac{-12(1)^2 + 110(1) - 86}{(1 - 4)^2} $$
$$ f'(1) = \frac{-12 + 110 - 86}{(-3)^2} $$
$$ f'(1) = \frac{12}{9} > 0 $$

Since $ f'(1) > 0 $, $ f(x) $ is increasing at $ x = 1 $.