Is #f(x)=(4x^3-2x^2-x-3)/(x-2)# increasing or decreasing at #x=0#?

Question

Emma Aldridge · Accepted Answer

increasing at x = 0

We evaluate f'(a) to find out if a function is increasing or decreasing at x = a. • At x = a, f(x) is increasing if f'(a) > 0. • If f'(a) < 0 , then f(x) is decreasing at x = a #"----------------------------------------------------------------------"# differentiate f(x) 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)#=4x^3-2x^2-x-3 rArr g'(x) = 12x^2 -4x - 1 # and h(x) = x-2 #rArr h'(x) = 1 # #"---------------------------------------------------------------------"# substitute these values into f'(x) f'(x) #= ((x-2)(12x^2-4x-1) - (4x^3-2x^2-x-3).1)/(x-2)^2 # and f'(0)# = ((-2)(-1) - (-3))/(-2)^2 = 5/4 # F(x) is increasing at x = 0 because f'(0) > 0. The graph {(4x^3-2x^2-x-3)/(x-2) [-10, 10, -5, 5]} shows this.

Ezra Adams · Answer

undefined To determine if $ f(x) = \frac{{4x^3 - 2x^2 - x - 3}}{{x - 2}} $ is increasing or decreasing at $ x = 0 $, we can evaluate the sign of the derivative of $ f(x) $ at that point.

First, find the derivative of $ f(x) $ using the quotient rule:

$$ f'(x) = \frac{{(x - 2)(12x^2 - 4x - 1) - (4x^3 - 2x^2 - x - 3)(1)}}{{(x - 2)^2}} $$

Now, evaluate $ f'(0) $ to determine if the function is increasing or decreasing at $ x = 0 $. If $ f'(0) > 0 $, then the function is increasing at $ x = 0 $; if $ f'(0) < 0 $, then the function is decreasing at $ x = 0 $.

$$ f'(0) = \frac{{-2(-1) - (-3)(1)}}{{(-2)^2}} = \frac{{2 + 3}}{{4}} = \frac{5}{4} $$

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