Determine if a function is strictly increasing/decreasing

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tompenny
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Homework Statement
Determine where a function is strictly increasing/decreasing
Relevant Equations
$$f(x)=x+\frac{1}{(x+1)}$$
Hi there.

I have the following function:

$$f(x)=x+\frac{1}{(x+1)}$$

I've caculated the derivative to:

$$f'(x)=1-\frac{1}{(1+x)^2}$$

And the domain to: $$(-\infty, -1)\cup(-1, \infty)$$

I've also found two extreme point: $$x=0, x=-2$$

I know that a function is strictly increasing if:
$$f'(x)> 0$$
and strictly decreasing if:
$$f'(x)< 0$$

I've calculated the intervalls where the funtcion is strictly increasing to:
$$(-\infty, -2]\cup[0, \infty)$$
and strictly decreasing to:
$$[-2, -1)\cup(-1, 0]$$

My question is if this is correct or if the intervalls should be:
$$(-\infty, -2)\cup(0, \infty)$$ and $$(-2, -1)\cup(-1, 0)$$ instead?
As you can notice I'm very unsecure whether I should use ( or [ at the extremum points?

Any help would be greatly appreciated.
Thank you:)

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  • #2
Can a function be strictly increasing and strictly decreasing at the same point?
 
  • #3
I really don't know.. are you reffering to the extremum point at x=-2?
 
  • #4
tompenny said:
I really don't know.. are you reffering to the extremum point at x=-2?
Yes. In your first solution you have the two points where ##f'(x) = 0## (i.e. ##x = 0## and ## x = -2##) as both strictly incraesing and strictly decreasing. That makes no sense.

Your second solution has got to be right.
 
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  • #5
PeroK said:
Can a function be strictly increasing and strictly decreasing at the same point?

It depends on what you mean by "strictly increasing at [itex]x[/itex]".

A function is strictly increasing or decreasing on some subset of its domain. It can't be strictly increasing or decreasing on a subset consisting of a single point, because the definitions don't really make sense in that case.[1]

Here we have:

It is true that [itex]f[/itex] is strictly increasing on [itex][0,\infty)[/itex].
It is true that [itex]f[/itex] is strictly decreasing on [itex](-1,0][/itex].

On the other hand, if by "[itex]f[/itex] is strictly increasing at [itex]x[/itex]" you mean "there is an open neighbourhood of [itex]x[/itex] on which [itex]f[/itex] is strictly increasing" then [itex]f[/itex] is not strictly increasing or decreasing at 0.

[1]: If [itex]X[/itex] consists of a single point then [tex](\forall x \in X)(\forall y \in X) ((x < y) \implies (f(x) < f(y)))[/tex] and [tex](\forall x \in X)(\forall y \in X) ((x < y) \implies (f(x) > f(y)))[/tex] are true, because whatever the choice of [itex]x[/itex] and [itex]y[/itex] the implications reduce to "false implies false", which is a true statement.
 
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