Question

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

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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`

Add the current column, and repeat the previous steps:

`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)`

This is indeed easier to differentiate, so now:

`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`.