Intervals of Increase and Decrease Explained: In-Depth Examples, Equations, and Expert Insights

Some questons about recurrence sequences (using a problem).
A quick doubt, lets study the recurrence sequence:
$A_{n+1}=(4A_n+2)/A_n+3$
$A_0<-<!--\; -\; -->3$
First of all i do:
$A_{n+1}-<!--\; -\; -->A_n<0$
If this is true i can say that $A_n$ decrease. This is true for those $A_n$ values:
$-<!--\; -\; -->3>A_n>-<!--\; -\; -->1$
$A_n>2$
And false (so $A_n$ increase) for those $A_n$ values:
$A_n<-<!--\; -\; -->3$
$-<!--\; -\; -->1<A_n<2$
the case $A_n<-<!--\; -\; -->3$ interests me.
The limit L can be -1 or -2 but i cannot say it exists for sure because $A_n$ is not limited and monotone for all the $A_n$ possible values. For example, the sequence can go from $A_n>2$ then decrease and go in $A_n<-<!--\; -\; -->3$ then again increase and fall in $A_n>2$ etc...
Another doubt comes from this fact:
It's ok to remove -1 from the possible values of L because in this case the sequence still growing?
Anyways: It happens so many times that i know the sequence increase or decrease in an interval but i don't know if doing it it will fall in another interval where it starts decreasing or increasing and in this scenario i don't know how to demonstrate if it goes on some limit or just starts to "ping-pong" on different intervals.
Or in other words i don't know how many it decrease/increase so i cannot say if it will go out from the interval i'm considering.

Trigonometric equation with two functions and a parameter
The problem: For which values of a the equation $a.sinx.cosx=sinx-<!--\; -\; -->cosx$ has exactly 2 different roots in the interval $[0;\mathrm{\pi <!--\; \pi \; -->}]$.
So clearly $0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2},\mathrm{\pi <!--\; \pi \; -->}$ are not an answer. I divide by sinx.cosx and get $f(x)=\frac{1}{cosx}-<!--\; -\; -->\frac{1}{sinx}=a$. I look at the interval $(0;\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$ cos x is decreasing so $\frac{1}{cosx}$ is increasing same for sin x is increasing and $\frac{-<!--\; -\; -->1}{sinx}$ is also increasing.
${\displaystyle \underset{x\to <!--\; \to \; -->0}{lim}f(x)=+\mathrm{\infty <!--\; \infty \; -->}}$ and ${\displaystyle \underset{x\to <!--\; \to \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}{lim}f(x)=-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}$ and f(x) is continuous so we have exactly 1 root in the interval $(0;\frac{\mathrm{\pi <!--\; \pi \; -->}}{2})$ for every a.
My question: I guess I have to find for what value of a the function has 1 root in the interval $(\frac{\mathrm{\pi <!--\; \pi \; -->}}{2};\mathrm{\pi <!--\; \pi \; -->})$ how do I do that and is there an easier way to solve the whole problem?

What is the critical point of this function?
The problem reads: $f(x)=7\frac{e^{2x}}{x}+4.$.
I am unsure of how to approach this problem to find the derivative. If someone could break down the steps that would be greatly appreciated.
Also, the question asks for intervals of the increasing and decreasing parts of the function. How would I figure this out? I'm thinking I'd use a sign chart. But if you have any other useful methods, I am all ears, or rather eyes.

Show the function is decreasing in an interval
Let $s>1,s\in <!--\; \in \; -->\mathbb{R}$, and let f be a function defined by
$f(x)=\frac{ln(x)}{x^s},x>0$
Determine the monotone intervals of f.
I note that f(x)'s domain is $(0,\mathrm{\infty <!--\; \infty \; -->})$. I then find the derivative of f(x).
$f^{\prime }(x)=x^{-<!--\; -\; -->s-<!--\; -\; -->1}(1-<!--\; -\; -->s\cdot <!--\; \cdot \; -->ln(x))$
$f^{\prime }(x)=0⟹<!--\; ⟹\; -->x=e^{\frac{1}{s}}$
So $e^{\frac{1}{s}}$ is a critical point for f.
I'm uncertain how to evaluate the function around the critical point. I'm thinking to evaluate at the points $e^{\frac{0}{s}}$ and $e^{\frac{2}{s}}$?
So $f^{\prime }(e^{\frac{0}{s}})=e^{{\frac{0}{s}}^{-<!--\; -\; -->s-<!--\; -\; -->1}}(1-<!--\; -\; -->s\cdot <!--\; \cdot \; -->ln(e^{\frac{0}{s}}))$.
I then need to figure out if the expression above is greater, less or equal to 0. And likewise with the other critical point.
I'm not sure if I've done this correctly and I'm not sure how to make the expression for the evaluated critical point any nicer.

Is it proper to say "increases/decreases from no bound"?
We commonly use the expression increases without bound to describe certain divergent behaviour of functions (e.g. the function $f(x)=x^2$ increases without bound on $[0,\mathrm{\infty <!--\; \infty \; -->})$). What would be the proper way of describing the behaviour of the curve of f(x) if I want to move towards the right on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]$?
In the curve sketching unit in my calculus course, I tend to describe all key elements (critical numbers intervals of increase/decrease, points of inflection, intercepts, etc.) from left to right on the x-axis, and so to increase consistency, I want to describe the increase/decrease of a function from left to right. So far, I have been saying "the function increases/decreases without bound towards the left", but this has been directly opposite to what that interval of the function is labelled. I want to know if there is better or more accurate language I could use.

Monotonicity of vector fields
If I have a vector field (x,y,z) on a one dimensional line (x-axis) and if I have to check its monotonicity between two intervals. Will it be monotonic if:
(1) only one of the components of vector field is monotonic in that interval. OR (2) two of the components of vector field are monotonicaly increasing and the other is monotonicaly decreasing in that interval. OR (3) all the components of vector field are either monotonicaly increasing or monotonicaly decreasing in that interval. .

If h has positive derivative and $\mathrm{\phi <!--\; \phi \; -->}$ is continuous and positive. Where is increasing and decreasing f
The problem goes specifically like this:
If h is differentiable and has positive derivative that pass through (0,0), and $\mathrm{\phi <!--\; \phi \; -->}$ is continuous and positive. If:
$f(x)=h({\int <!--\; \int \; -->}_0^{\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}}\mathrm{\phi <!--\; \phi \; -->}(t)dt).$
Find the intervals where f is decreasing and increasing, maxima and minima.
My try was this:
The derivative of f is given by the chain rule:
$f^{\prime }(x)=h^{\prime }({\int <!--\; \int \; -->}_0^{\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}}\mathrm{\phi <!--\; \phi \; -->}(t)dt)\mathrm{\phi <!--\; \phi \; -->}(\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2})$
We need to analyze where is positive and negative. So I solved the inequalities:
$\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}>0\wedge <!--\; \wedge \; -->\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}<0$
That gives: $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2)\cup <!--\; \cup \; -->(\surd 2,\mathrm{\infty <!--\; \infty \; -->})$ for the first case and $(-<!--\; -\; -->\surd 2,\surd 2)$ for the second one. Then (not sure of this part) $h^{\prime }({\int <!--\; \int \; -->}_0^{\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2}}\mathrm{\phi <!--\; \phi \; -->}(t)dt)>0$ and $\mathrm{\phi <!--\; \phi \; -->}(\frac{x^4}{4}-<!--\; -\; -->\frac{x^2}{2})>0$ if $x\in <!--\; \in \; -->(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2)\cup <!--\; \cup \; -->(\surd 2,\mathrm{\infty <!--\; \infty \; -->}).$ Also if both h′ and $\mathrm{\phi <!--\; \phi \; -->}$ are negative the product is positive, that's for $x\in <!--\; \in \; -->(-<!--\; -\; -->\surd 2,\surd 2)$.
The case of the product being negative implies:
$x\in <!--\; \in \; -->[(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2)\cup <!--\; \cup \; -->(\surd 2,\mathrm{\infty <!--\; \infty \; -->})]\cap <!--\; \cap \; -->(-<!--\; -\; -->\surd 2,\surd 2)=[(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2)\cap <!--\; \cap \; -->(-<!--\; -\; -->\surd 2,\surd 2)]\cup <!--\; \cup \; -->[(\surd 2,\mathrm{\infty <!--\; \infty \; -->})\cap <!--\; \cap \; -->(-<!--\; -\; -->\surd 2,\surd 2)]=\mathrm{\varnothing <!--\; \varnothing \; -->}.$
So the function is increasing in $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\surd 2),(-<!--\; -\; -->\surd 2,\surd 2),(\surd 2,\mathrm{\infty <!--\; \infty \; -->})$. So the function does not have maximum or minimum. Not sure of this but what do you think?

Determine the monotonic intervals of the functions
i need to determine the monotonic intervals of this function $y=2x^3-<!--\; -\; -->6x^2-<!--\; -\; -->18x-<!--\; -\; -->7$. I tried the below but i am not sure if i am doing it right.
My work: $\begin{array}{rl}y=2x^3-<!--\; -\; -->6x^2-<!--\; -\; -->18x-<!--\; -\; -->7& ⟺<!--\; ⟺\; -->6x^2-<!--\; -\; -->12x-<!--\; -\; -->18=0\\ & ⟺<!--\; ⟺\; -->6(x^2-<!--\; -\; -->2x-<!--\; -\; -->3)=0\\ & ⟺<!--\; ⟺\; -->(x-<!--\; -\; -->3)(x+1)\\ & ⟺<!--\; ⟺\; -->x-<!--\; -\; -->3=0x+1=0\\ & ⟺<!--\; ⟺\; -->x=3,x=-<!--\; -\; -->1\end{array}$
so my function increases when $x\in <!--\; \in \; -->[3,+\mathrm{\infty <!--\; \infty \; -->}[$ and decreases when $x\in <!--\; \in \; -->[-<!--\; -\; -->1,3]\cup <!--\; \cup \; -->]-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->1]$.
Please i want to know how to solve this problem any help with explanation will be appreciated. thanks in advanced

The effect of changing sample sizes on outliers
I know that the size of a sample is inversely proportional to the width of a confidence interval, and that outliers tend to increase the width of the interval as well. So that must mean that increasing the sample size reduces the effect of outliers on a confidence interval, and decreasing the sample size amplifies the effect, correct?
How can I show this using formulae instead of words for, say a confidence interval for a one-sample z-test?
Also as a side note, does changing the sample size change how outliers affect the p-value of a hypothesis test? I'm inclined to say yes, but I'm not sure how to justify that conclusion.

Confusion between the definition of increasing function and a theorem regarding it.
The definition of increasing function given in my school maths text book is
Let I be an open interval contained in the domain of a real valued function f. Then f is said to be increasing on I if $a<<!--\; <\; -->b⟹<!--\; ⟹\; -->f(a)\le <!--\; \le \; -->f(b)$ for all $a,b\in <!--\; \in \; -->I$.
And a theorem given after it is
f is increasing in I if $f^{\prime }(x)><!--\; >\; -->0\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->I$.
Shouldn't it be $f^{\prime }(x)\ge <!--\; \ge \; -->0$.
Is constant function an increasing function? Or a function like $f(x)=x^3$ where $f^{\prime }(x)=0$ at some or all points which are increasing according to definition.
Edit: The definition of decreasing function given is
f is decreasing on I if $a<<!--\; <\; -->b⟹<!--\; ⟹\; -->f(a)\ge <!--\; \ge \; -->f(b)$.
A constant function follow this definition too.

Sketch $f(x)=\sin <!--\; \; -->x+\frac{1}{x}$ finding local maxima and minima, intervals of increase and decrease. I'm trying to use differentiation to draw this picture and find critical points.
So, I get $f^{\prime }(x)=\cos <!--\; \; -->x-<!--\; -\; -->\frac{1}{x^2}$. However, I'll have to deal with inequality $f^{\prime }(x)>0$, and $f^{\prime }(x)=0$, I feel I lack an ability to solve an equation like this. So is there any better way to find local maximun and minumum for this function?

Increasing and decreasing intervals
Find a polynomial f(x) of degree 4 which increases in the intervals $(-<!--\; -\; -->\infty ,1)$ and (2,3) and decreases in the intervals (1,2) and $(3,\infty )$ and satisfies the condition $f(0)=1$.
It is evident that the function should be $f(x)=a{x}^4+b{x}^3+c{x}^2+dx+1$. I differentiated it. Now, I'm lost. I tried putting $f^{\prime }(1)>0$, $f^{\prime }(3)-<!--\; -\; -->f^{\prime }(2)\ge <!--\; \ge \; -->0$, and $f^{\prime }(2)-<!--\; -\; -->f^{\prime }(1)\le <!--\; \le \; -->0$. Am I doing correct? Or is there another method?

Find interval of increase and decrease
So im supposed to find the interval of decrease and increase here. Ive gotten up to taking the derivative which is $-<!--\; -\; -->4x(x^2-<!--\; -\; -->1)$ and then setting it to 0 i got (-1,0,1) Im lost at what to do now?
Im supposed to take it for this below:
$f(x)=7+2x^2-x^4$

Question about strictly increasing and continuous functions on an interval
If $F,\mathrm{\phi <!--\; \phi \; -->}:[a,b]\to <!--\; \to \; -->\mathbb{R}$ with F continuous, $\mathrm{\phi <!--\; \phi \; -->}$ strictly increasing and $(F+\mathrm{\phi <!--\; \phi \; -->})(a)>(F+\mathrm{\phi <!--\; \phi \; -->})(b)$ is $F+\mathrm{\phi <!--\; \phi \; -->}$ decreasing on some interval in (a,b)?
This is clearly true if $\mathrm{\phi <!--\; \phi \; -->}$ has finitely many points of discontinuity in the interval but I am unsure if this statement is true if there are infinitely many points of discontinuity.

Find the Intervals of Increase or Decrease, the concavity, and point of inflection for: $f(x)=(1-<!--\; -\; -->x)e^{-<!--\; -\; -->x}$.
$\begin{array}{rl}f(x)& =(1-<!--\; -\; -->x)e^{-<!--\; -\; -->x}\\ & =\frac{1-<!--\; -\; -->x}{e^x}\end{array}$
Quotient Rule:
$\begin{array}{rl}{f}^{\prime }(x)& =e^x\cdot <!--\; \cdot \; -->[1-<!--\; -\; -->x{]}^{\prime }-<!--\; -\; -->(1-<!--\; -\; -->x)\cdot <!--\; \cdot \; -->[e^x{]}^{\prime }\\ & =e^x(-<!--\; -\; -->2+x)\end{array}$
What do I do now? Take the derivative again?

Is $f(x)=x\sqrt{4-<!--\; -\; -->x}$ decreasing at $x=4$?
I am solving a single variable calculus problem and it asks me to determine the decreasing and decreasing intervals of the function $f(x)=x\sqrt{4-<!--\; -\; -->x}$. It's pretty clear that from $]-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\frac{8}{3}[$ the function is increasing but, since the domain of the function goes only until $x=4$, should I write that the decreasing interval of the function is $]\frac{8}{3},4[$ or $]\frac{8}{3},4]$?

Figuring out when $f(x)=\sin <!--\; \; -->(x^2)$ is increasing and decreasing
Regarding the function $f(x)=\sin <!--\; \; -->(x^2)$, I'm supposed to figure out when it is increasing/decreasing.
So far, I've found the derivative to be $f^{\prime }(x)=2x\cos <!--\; \; -->(x^2)$.
So long as I can solve the inequality $\cos <!--\; \; -->(x^2)>0$. I can figure the rest out, but this is where I'm stuck.
I've narrowed it down to $-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}<x^2<\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$ meaning $x^2<\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}\pm <!--\; \pm \; -->n2\mathrm{\pi <!--\; \pi \; -->},n\in <!--\; \in \; -->\mathbb{N}$.
Then I get $x<\pm <!--\; \pm \; -->\sqrt{\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}\pm <!--\; \pm \; -->n2\mathrm{\pi <!--\; \pi \; -->}}$.
Am I on the right track here? I can't seem to find a path from here.

Graphing using derivatives
Sketch the graph of the following equation. Show steps of finding out critical numbers, intervals of increase and decrease, absolute maximum and minimum values and concavity.
$y=x{e}^{x^2}$
I found the first derivative which is $y^{\prime }=(2x^2+1)e^{x^2}$ and I know that in order to find min and max the zeroes for y′ must be found, but y′ doesn't have any real zeroes, and I'm confused about how to go on with solving the problem.
If someone could help me out, that would be appreciated. Thank you in advance.

Is a Square Bracket Used in Intervals of Increase/Decrease?
For example, the I.O.I of $y=x^2$ is (0,infinite), with the round brackets meaning that the value is excluded. Are there any scenarios where a square bracket would be used when stating the intervals of increase/ decrease for a function? If it narrows it down, the only functions I deal with are: linear, exponential, quadratic, root, reciprocal, sinusoidal, and absolute

Find the rate of change of main dependent variable
We have $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R},f(x)=x^2+x\sin <!--\; \; -->(x)$, and we need to find intervals of monotonicity. Here is all my steps:
$f^{\prime }(x)=2x+x\cos <!--\; \; -->(x)+\sin <!--\; \; -->(x)$
$f^{\prime }(x)=0\Rightarrow <!--\; \Rightarrow \; -->x=0$ the only solution.
Now I need to find where f′ is positive and negative. I don't want to put value for f′ to find the sign, I want another method. So I tried to differentiate the function again to see if f′ is increasing or decreasing:
$f^{″}(x)=2-<!--\; -\; -->x\sin <!--\; \; -->(x)+\cos <!--\; \; -->(x)$, but I don't know if $f^{″}(x)\ge <!--\; \ge \; -->0$ or $f^{″}(x)\le <!--\; \le \; -->0$.
How can I find the sign for f′ to determine monotony of f ?