Is #f(x)=(-2x^3+9x^2-5x+6)/(x-2)# increasing or decreasing at #x=0#?

Question

Isaiah Allen · Accepted Answer

#f# is #uarr# at #x=0#.

We know that, #f# is #uarr or darr at x=0# according as f'(0) >, or, < 0"#. So, we need to check #f'(0)#. Now, #f(x)=(-2x^3+9x^2-5x+6)/(x-2)# #={ul(-2x^3+4x^2)+ul(5x^2-10x)+ul(5x-10)+16}/(x-2)# #={-2x^2(x-2)+5x(x-2)+5(x-2)+16}/(x-2)# #={(cancel((x-2))(-2x^2+5x+5))/cancel((x-2))+16/(x-2)}# #:. f(x)=-2x^2+5x+5+16/(x-2)................(1)# Knowing that, #d/dt(1/t)=-1/t^2, "we have, by" (1)#, #f'(x)=-4x+5-16/(x-2)^2# #rArr f'(0)=5-16/(-2)^2=5-4=1>0# Hence, #f# is #uarr# at #x=0#. Have fun with math!

Charles Anctil · Answer

undefined To determine whether the function $ f(x) = \frac{{-2x^3 + 9x^2 - 5x + 6}}{{x - 2}} $ is increasing or decreasing at $ x = 0 $, we can analyze its derivative at that point.

1. First, find the derivative of the function $ f(x) $.
2. Evaluate the derivative at $ x = 0 $.
3. Determine the sign of the derivative at $ x = 0 $.

The function is increasing at $ x = 0 $ if the derivative is positive, and it's decreasing if the derivative is negative.

Let's find the derivative:

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

Using the quotient rule to differentiate, we get:

$ f'(x) = \frac{{(x - 2)(-6x^2 + 18x - 5) - (-2x^3 + 9x^2 - 5x + 6)(1)}}{{(x - 2)^2}} $

Now, evaluate $ f'(0) $ to determine whether the function is increasing or decreasing at $ x = 0 $.

$ f'(0) = \frac{{(-2)(-5) - (6)(1)}}{{(-2)^2}} = \frac{{10 - 6}}{{4}} = \frac{4}{4} = 1 $

Since $ f'(0) = 1 > 0 $, the function $ f(x) $ is increasing at $ x = 0 $.