# Determine if a function is strictly increasing/decreasing

tompenny
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:)

http://asciimath.org/
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Can a function be strictly increasing and strictly decreasing at the same point?

tompenny
I really don't know.. are you reffering to the extremum point at x=-2?

Homework Helper
Gold Member
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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.

• tompenny
Homework Helper
2022 Award
Can a function be strictly increasing and strictly decreasing at the same point?

It depends on what you mean by "strictly increasing at $x$".

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.

Here we have:

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

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

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

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