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

1 Answer
Dec 5, 2017

Decreasing; see explanation

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]}