Is #f(x)=(x^3+3x^2-4x-9)/(x+1)# increasing or decreasing at #x=0#?

Answer 1
It's increasing, because the slope is positive at #x = 0#.

graph{[-5, 5, -12, 5]}/(x+1) = (x^3 + 3x^2 - 4x - 9)

The first derivative tells you whether the function is decreasing (#(df)/(dx) < 0#), increasing (#(df)/(dx) > 0#), or at an extremum (a turning point, #(df)/(dx) = 0#).
For this, I would actually try to do some quick synthetic division to simplify it if possible. I would guess #x = -1# in hopes of cancelling out the denominator, but if it doesn't work, that's OK. It'll at least become simpler.
#ul(-1)|" "1" "3" "-4" "-9# #"-----------------------------------"# #" "" "" "1#
Multiply the divisor (the factor you guessed, #-1#) by the term you just brought down (#1#), and place it under the next column (#-1xx1 = -1#).
#ul(-1)|" "1" "3" "-4" "-9# #" "" "" "" "-1# #"-----------------------------------"# #" "" "" "1#

Repeat the preceding actions and add the current column:

#ul(-1)|" "1" "3" "-4" "-9# #" "" "" "" "-1# #"-----------------------------------"# #" "" "" "1" "2#
#ul(-1)|" "1" "3" "-4" "-9# #" "" "" "" "-1" "-2# #"-----------------------------------"# #" "" "" "1" "2" "-6#
#ul(-1)|" "1" "3" "-4" "-9# #" "" "" "" "-1" "-2" "" "6# #"-----------------------------------"# #" "" "" "1" "2" "-6" "-3#
You started with a cubic, so you end up with a quadratic. The remainder is #-3#, and one way of writing it is to leave it divided by #x+1# (your denominator):
#x^2 + 2x - 6 - 3/(x+1)#

It is now simpler to distinguish between these, so

#d/(dx)[x^2 + 2x - 6 - 3/(x+1)]#
#= 2x + 2 + 3/(x+1)^2#
Now, plug in #x = 0# to find out what is going on with this function at #x = 0#:
#f'(0) = |[(dy)/(dx)]|_(x=0) = 2(0) + 2 + 3/(0+1)^2 = 5 > 0#
Therefore, the function is increasing at #x = 0#.
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Answer 2

To determine whether ( f(x) = \frac{x^3 + 3x^2 - 4x - 9}{x + 1} ) is increasing or decreasing at ( x = 0 ), we can use the first derivative test. First, find the derivative of ( f(x) ), then evaluate the derivative at ( x = 0 ). If the derivative is positive at ( x = 0 ), the function is increasing at that point. If the derivative is negative, the function is decreasing.

( f'(x) = \frac{d}{dx}\left(\frac{x^3 + 3x^2 - 4x - 9}{x + 1}\right) )

Using the quotient rule:

( f'(x) = \frac{(x+1)(3x^2 + 6x - 4) - (x^3 + 3x^2 - 4x - 9)(1)}{(x + 1)^2} )

Simplify and evaluate ( f'(0) ):

( f'(0) = \frac{(0 + 1)(3(0)^2 + 6(0) - 4) - ((0)^3 + 3(0)^2 - 4(0) - 9)(1)}{(0 + 1)^2} )

( f'(0) = \frac{(1)(-4) - (-9)(1)}{(1)^2} )

( f'(0) = \frac{-4 + 9}{1} )

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

Since ( f'(0) = 5 > 0 ), the function is increasing at ( x = 0 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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