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

Question

Aurora Alvarado · Accepted Answer

Decreasing; see explanation

In order to determine the solution, we must find the derivative #f'(3)# The Quotient Rule states that, for #f(x) =(g(x))/(h(x)), f'(x) = (g'(x)h(x) - g(x)h'(x))/(h^2(x))# In this case, we have via the power rule: #g(x) = 3x^3+3x^2+5x+2, g'(x) = 9x^2+6x + 5, h(x) = x-2, h'(x) = 1# Thus: #f'(x) = ((9x^2+6x+5)(x-2) - (3x^3+3x^2+5x+2))/(x-2)^2# We can simplify further if we wish, but for the purposes of this problem it's unnecessary. Now find #f'(c)# by plugging in c... #f'(3) = ((9(3^2)+6(3) + 5)(3-2) - (3(3^3) + 3(3^2)+5(3)+2))/((3-2)^2)# #= ((81+18+5)(1) - (81+27+15+2))/(1^2) = -21/1 = -21# Because the derivative is negative at this point, #f(x)# is decreasing at #x=3# Viewing the graph below, we can verify this is correct. graph{(3x^3 + 3x^2 + 5x + 2)/(x-2) [-2.947, 9.2, 121.787, 127.86]}

Emma Aldridge · Answer

undefined To determine if $ f(x) = \frac{3x^3 + 3x^2 + 5x + 2}{x - 2} $ is increasing or decreasing at $ x = 3 $, we can use the first derivative test. If the derivative is positive at $ x = 3 $, then $ f(x) $ is increasing at that point. If the derivative is negative at $ x = 3 $, then $ f(x) $ is decreasing at that point.

The derivative of $ f(x) $ is $ f'(x) = \frac{d}{dx}\left(\frac{3x^3 + 3x^2 + 5x + 2}{x - 2}\right) $.

Using the quotient rule, we find:

$ f'(x) = \frac{(x - 2)(9x^2 + 6x + 5) - (3x^3 + 3x^2 + 5x + 2)(1)}{(x - 2)^2} $.

Evaluating this at $ x = 3 $, we get:

$ f'(3) = \frac{(3 - 2)(9(3)^2 + 6(3) + 5) - (3(3)^3 + 3(3)^2 + 5(3) + 2)}{(3 - 2)^2} $.

Simplifying:

$ f'(3) = \frac{(1)(81 + 18 + 5) - (27 + 9 + 15 + 2)}{1} $,

$ f'(3) = \frac{104 - 53}{1} $,

$ f'(3) = \frac{51}{1} $,

$ f'(3) = 51 $.

Since $ f'(3) = 51 $ is positive, $ f(x) $ is increasing at $ x = 3 $.