Is #f(x)=(-2x^3-x^2-5x+2)/(x+1)# increasing or decreasing at #x=0#?

Question

Christopher Agresta · Accepted Answer

decreasing at x = 0

Verify the sign of f'(a) in order to determine whether a function is increasing or decreasing at x = a. • At x = a, f(x) is increasing if f'(a) > 0. • At x = a, f(x) is decreasing if f'(a) < 0. Need to determine f'(x) differentiate using the #color(blue)" Quotient rule " # If # f(x) = g(x)/(h(x)) " then " f'(x) = (h(x).g'(x) - g(x).h'(x))/(h(x))^2 # #"-------------------------------------------------------------------------"# #g(x) = -2x^3 - x^2- 5x + 2 rArr g'(x) =-6x^2-2x-5# h(x) = x+1 # rArr h'(x) = 1# #"------------------------------------------------------------------------"# substitute these values into f'(x) #rArr f'(x) =( (x+1).(-6x^2-2x-5) - (-2x^3-x^2-5x+2).1)/(x+1)^2# and f'(0) = #(1.(-5) - 2.1)/1 = -7 # At x = 0, f(x) is decreasing because f'(0) < 0. The graph representing f(x) is as follows: graph{(-2x^3-x^2-5x+2)/(x+1) [-10, 10, -5, 5]}

Evelyn Agard · Answer

undefined To determine whether the function $ f(x) = \frac{-2x^3 - x^2 - 5x + 2}{x + 1} $ is increasing or decreasing at $ x = 0 $, we can use the first derivative test.

First, find the first derivative of the function:
$$ f'(x) = \frac{d}{dx} \left( \frac{-2x^3 - x^2 - 5x + 2}{x + 1} \right) $$

After finding the derivative, evaluate it at $ x = 0 $. If the derivative is positive, the function is increasing at that point. If it's negative, the function is decreasing.

$$ f'(0) = \frac{d}{dx} \left( \frac{-2x^3 - x^2 - 5x + 2}{x + 1} \right) \Bigg|_{x=0} $$

$$ f'(0) = \frac{(-6x^2 - 2x - 5)(x + 1) - (-2x^3 - x^2 - 5x + 2)(1)}{(x + 1)^2} \Bigg|_{x=0} $$

$$ f'(0) = \frac{(-5)(1) - (2)(1)}{(1)^2} $$

$$ f'(0) = \frac{-5 - 2}{1} $$

$$ f'(0) = -7 $$

Since the derivative at $ x = 0 $ is negative ($ f'(0) = -7 $), the function is decreasing at that point.