Is #f(x)=(-x^2-5x-2)/(x^2+1)# increasing or decreasing at #x=-3#?

Question

Isaiah Allen · Accepted Answer

Increasing.

Find the value of the derivative at #x=-3#. If it is positive, the function is increasing. If it's negative, the function is decreasing (at that point). Apply the quotient rule to determine the function's derivative. #f'(x)=((x^2+1)d/dx[-x^2-5x-2]-(-x^2-5x-2)d/dx[x^2+1])/(x^2+1)^2# The power rule can be used to find each of these derivatives: #d/dx[-x^2-5x-2]=-2x-5# #d/dx[x^2+1]=2x# Re-plugging these produces #f'(x)=((x^2+1)(-2x-5)+(x^2+5x+2)(2x))/(x^2+1)^2# Derivative of the numerator after distribution and simplification #f'(x)=(5x^2+2x-5)/(x^2+1)^2# We can now find the value of the derivative at #x=-3:# #f'(-3)=(5(9)-6-5)/(9+1)^2=34/100=17/50# Since this is #>0#, the function is increasing at #x=-3#. We can check the graph of #f(x):# graph{(x^2+1) [-5, 5, -5.89, 3]} / (x^2-5x-2)

Paisley Johnson · Answer

undefined To determine whether $ f(x) = \frac{-x^2 - 5x - 2}{x^2 + 1} $ is increasing or decreasing at $ x = -3 $, we need to evaluate the sign of the derivative of $ f(x) $ at that point.

1. Find the derivative of $ f(x) $ using the quotient rule:
$$ f'(x) = \frac{(x^2 + 1)(-2x - 5) - (-x^2 - 5x - 2)(2x)}{(x^2 + 1)^2} $$

2. Evaluate $ f'(-3) $:
$$ f'(-3) = \frac{((-3)^2 + 1)(-2(-3) - 5) - (-(-3)^2 - 5(-3) - 2)(2(-3))}{((-3)^2 + 1)^2} $$
$$ = \frac{(9 + 1)(6 - 5) - (9 + 15 - 2)(-6)}{(9 + 1)^2} $$
$$ = \frac{(10)(1) - (22)(-6)}{(10)^2} $$
$$ = \frac{10 + 132}{100} $$
$$ = \frac{142}{100} $$
$$ = 1.42 $$

Since $ f'(-3) = 1.42 > 0 $, $ f(x) $ is increasing at $ x = -3 $.