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

Non-decreasing sequences
I have the following sequence
$\{n^4-<!--\; -\; -->6n^2\}$
I have to determine if the sequence is non-decreasing, increasing or decreasing.
In my opinion, the sequence is neither, decreasing, non-decreasing nor increasing because it seems to be increasing for all the terms except for the first and second term. Am I right or I am right?
By the way,is there category for such a sequence? where you have an interval of increasing terms and one of decreasing terms or is such a sequence not even possible?

Calculus, solving for increasing/decreasing and concavity
Determine the intervals where the function is increasing and where it is decreasing. (Enter your answers using interval notation.)
$f(x)=\frac{\ln <!--\; \; -->(x)}{x}$
Find intervals of concavity for the graph of the function $f(x)=\frac{\ln <!--\; \; -->(x)}{x}$.
I have already found the first and second derivative but I am confused on how to solve for it $f^{\prime }(x)=-<!--\; -\; -->\frac{\ln <!--\; \; -->(x)-<!--\; -\; -->1}{x^2}$
$f^{″}(x)=\frac{2\ln <!--\; \; -->(x)-<!--\; -\; -->3}{x^3}$

Intervals of increase and decrease
If you have a function and there's an asymptote at say -7, then when doing the intervals for increase decrease, would you do something like increasing from $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->7)\cup <!--\; \cup \; -->(-<!--\; -\; -->7,\text{wherever increase stops})$ and not include the -7, or would the -7 be included

Why do I get different results when testing for increasing/decreasing intervals of a function here?
I have the function $f(x)=6+\frac{6}{x}+\frac{6}{x^2}$ and I want to find the intervals where it increases or decreases. The problem is that when I find $f^{\prime }(x)=0$, which becomes $x=-<!--\; -\; -->2$. Once I put -2 on a number line, and find whether the numbers higher and lower than -2 produce positive or negative numbers when plugged into f′(x), I get both positives and negatives for different numbers when $x>-<!--\; -\; -->2$. For example, $f^{\prime }(-<!--\; -\; -->1)>0$ and $f^{\prime }(5)<0$.
I am not sure if this is because the function has a vertical asymptote at $x=0$ and a horizontal attribute at $y=6$. If so, I'd like to know what I need to do to handle them.
How I found the first derivative:
$\frac{d}{dx}\frac{6}{x}=\frac{[0]-<!--\; -\; -->[6\cdot <!--\; \cdot \; -->1]}{x^2}=\frac{-<!--\; -\; -->6}{x^2}$
$\frac{d}{dx}\frac{6}{x^2}=\frac{[0]-<!--\; -\; -->[6\cdot <!--\; \cdot \; -->2x]}{(x^2{)}^2}=\frac{-<!--\; -\; -->12}{x^3}$
$f^{\prime }(x)=\frac{-<!--\; -\; -->6}{x^2}-<!--\; -\; -->\frac{-<!--\; -\; -->12}{x^3}$
How I found the asymptotes:
When you combine the fractions of f(x), $f(x)=\frac{6x^2+6x+6}{x^2}$. When set equal to zero, $x^2$ has an x-value of zero. Therefore, f(x) has a vertical asymptote at $x=0$.
Once the fractions are combined, and when everything but the coefficients of the leading terms on the numerator and denominator are removed, $f(x)=\frac{6}{1}$. So f(x) has a horizontal asymptote at $y=6$.
How I found the value of $f^{\prime }(x)=0$:
$\frac{-<!--\; -\; -->6}{x^2}-<!--\; -\; -->\frac{12}{x^3}=0$
Add $\frac{12}{x^3}$ to both sides
$\frac{6}{x^2}=\frac{-<!--\; -\; -->12}{x^3}$
Multiply both sides by $x^3$.
$6x=-<!--\; -\; -->12$
Divide both sides by 6.
$x=\frac{-<!--\; -\; -->12}{6}=-<!--\; -\; -->2$

Increasing and decreasing intervals of a function
$f(x)=x^3-<!--\; -\; -->4x^2+2$, which of the following statements are true:
(1) Increasing in $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty <!--\; \infty \; -->})$.
(2) Increasing in both $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)$, decreasing in $(\frac{8}{3},+\mathrm{\infty <!--\; \infty \; -->})$.
(3) decreasing in both $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)$, and $(\frac{8}{3},+\mathrm{\infty <!--\; \infty \; -->})$.
(4) Decreasing in $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)$, Increasing in $(\frac{8}{3},+\mathrm{\infty <!--\; \infty \; -->})$.
(5) None of the above.
$f^{\prime }(x)=0=3x^2-<!--\; -\; -->8x=0\Rightarrow <!--\; \Rightarrow \; -->x=\frac{8}{3},x=0$ are the singular point/point of inflection.Could anyone tell me what next?

Interval related to increasing/decreasing and concavity/convexity
Why do some people use closed intervals when describing the intervals where a function is increasing/decreasing or concave/convex?
For example, given the function $f(x)=x^2-<!--\; -\; -->5x+6$, it says the interval of increase is $[5/2,\mathrm{\infty <!--\; \infty \; -->})$. Why is this written as a closed interval, and not an open one?
Concavity, on the other hand, uses open intervals.

If $\tan <!--\; \; -->(x)>x$. Find interval of x.
We encountered this question in the class today. My lecturer asked me to define a function $f(x)=\tan <!--\; \; -->(x)-<!--\; -\; -->x$. Then we were asked to find its derivative, f′(x), i.e, ${\sec }^2<!--\; \; -->(x)-<!--\; -\; -->1$, and solve the inequality $f^{\prime }(x)>0$. The answer was $(0,\pi /2)$.
My question is, if $f^{\prime }(x)>0$, it just proves that f(x) is increasing in the interval, it doesn't prove if f(x) is always positive there. Ok, agreed that since $\tan <!--\; \; -->0=0$, $\tan <!--\; \; -->x>x$ in $(0,\pi /2)$ since it's increasing. But there is possibility of loosing a solution right? It could be possible that in an interval $\tan <!--\; \; -->x-<!--\; -\; -->x$ is decreasing but tan(x) is still $>x$

Increasing and decreasing piecewise function on an interval
I'm working on a problem that involves finding the intervals where a function f is increasing and decreasing. Given the function $f(x)=\{\begin{array}{ll}x+7& \text{ text{if } xlt -3}\\ |x+1|& \text{ text{if } -3le x <1}\\ 5-<!--\; -\; -->2x& \text{ text{if } xge 1}\end{array}$
I worked out that f is increasing on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->3)$ and $[-<!--\; -\; -->1,1)$, and f is decreasing on $[-<!--\; -\; -->3,-<!--\; -\; -->1],[1,\mathrm{\infty <!--\; \infty \; -->})$.
However, the solution in the book claims that f is increasing on [-1,1], rather than [-1,1) like I worked out. I am having trouble understanding why the book claims that this is the case.
I was under the impression that f must be differentiable on the interior of an interval I and continuous on all of I in order to make any statements about increasing/decreasing behavior on the closed interval I. I'm not sure if I am overlooking something, but it seems that f is not continuous on [-1,1]. I would appreciate any clarification.

Applying IV theorem to practical problems
Prove using the IVT: For all positive integers k, $cosx=x^k$ has a solution
I know that the IV states at an interval [a,b] some value c is bound to be hit atleast once, assuming that [a,b] is continuous(Please improve my explanation where needed).
So, using this logic, assuming that $0=x^k-<!--\; -\; -->cosx$, I have to find values of x where this would be true?
$f(0)=0-<!--\; -\; -->1$
$f(\pi /2)=(\pi /2{)}^k-<!--\; -\; -->0$
$f(\pi /2)>0$
By the IVT, there should be a value C inbetween $[0,\pi /2]$ if I am correct?
Please correct me where possible. I think I understand the theorem but applying it to a question like this makes me believe otherwise.

Value of f′(6) in given polynomial
A polynomial function f(x) of degree 5 with leading coefficients one , increase in the interval
$(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)$ and $(3,\mathrm{\infty <!--\; \infty \; -->})$ and decrease in the interval (1,3). Given that $f^{\prime }(2)=0,f(0)=4$.
Then value of $f^{\prime }(6)=$
Attempt: Given function increase in the interval $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)$ and $(3,\mathrm{\infty <!--\; \infty \; -->})$ and decrease
in the interval (1,3). So we have $f^{\prime }(1)=0$ and $f^{\prime }(3)=0$.
So $f^{\prime }(x)=(x-<!--\; -\; -->1)(x-<!--\; -\; -->2)(x-<!--\; -\; -->3)(x-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$.
Could some help me how i calculate $\mathrm{\alpha <!--\; \alpha \; -->},$

Find range of values
Find the range of values of the constant a at which the equation $x^3-<!--\; -\; -->3a^{2}x+2=0$ has 3 different real number roots.
I took the derivative and found that $x=-<!--\; -\; -->a,a$. Then I solved for $f(a)=0$ and $f(-<!--\; -\; -->a)=0$ to find that $a=-<!--\; -\; -->1,1$.
How do I use this information to find the range of values, or am I on the wrong path completely?

For any $n^2+1$ closed intervals of $\mathbb{R}$, prove that $n+1$ of the intervals share a point or $n+1$ of the intervals are disjoint
Stuck on a question from 'Introduction to Combinatorics by Martin J. Erickson'.
Q: For any $n^2+1$ closed intervals of $\mathbb{R}$, prove that $n+1$ of the intervals share a point or $n+1$ of the intervals are disjoint.
I think we can use the Erdos-Szekeres Theorem: relating the usual series of integers to the closed intervals, and the decreasing/increasing monotonic sequences somehow to the two outcomes, but I am stuck on how to technically do this.
Could we measure the 'distance' between the intervals? Creating a decreasing sequence ensuring they'd be close enough to share a point, and an increasing sequence that would ensure they are far enough away to be disjoint?

Find intervals decrease and increase of a function $f(x)=\frac{\mathrm{\pi <!--\; \pi \; -->}x}{2}-<!--\; -\; -->x\arctan <!--\; \; -->x$.
Find intervals decrease and increase of a function.
$f(x)=\frac{\mathrm{\pi <!--\; \pi \; -->}x}{2}-<!--\; -\; -->x\arctan <!--\; \; -->x$
$f^{\prime }(x)=\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->\arctan <!--\; \; -->x-<!--\; -\; -->\frac{x}{1+x^2}$
$f^{″}(x)=\frac{-<!--\; -\; -->2}{(1+x^2{)}^2}$
which is negative. What can I say from here?

Fourth solution of $y^{\prime }={\displaystyle \frac{10}{3}}x{y}^{2/5}$, $y(0)=0$?
By inspection, we know that $y=0$ is a solution. If we separate variables, we get another solution, $y=x^{10/3}$. The book tells me this much, and asks me to find the other 2 solutions that differ on every open interval containing $x=0$ and are defined on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\mathrm{\infty <!--\; \infty \; -->})$.
I found the third solution (by guessing) to be $y=-<!--\; -\; -->x^{10/3}$. The only other possibility i could think about is $x=0$, but I'm pretty sure that's not valid since y′ would be undefined. What is the last solution? How do we know there are only 4 possible solutions (and not more)?

Let $f(x)=\frac{\ln <!--\; \; -->(x)}{\sqrt{x}}$. Find the critical values of f(x)
How would you find the critical value of this equation? So far... I have gotten that you find the derivative of f(x) which is $f^{\prime }(x)=\frac{1}{x^{3/2}}-<!--\; -\; -->\frac{\ln <!--\; \; -->(x)}{2x^{3/2}}.$.
To find the critical values set $f^{\prime }(x)=0$? But how would you factor out the derivative to find the critical values?
Also, after that how would you find the intervals of increase and decrease and relative extreme values of f(x).
And lastly the intervals of concavity and inflection points.

Finding the critical points of a function and the interval where it increases and decreases.
I am having trouble finding the critical points of
$f(x)=(x+1)/x-<!--\; -\; -->3$
I found the derivative to be $f^{\prime }(x)=-<!--\; -\; -->4/(x-<!--\; -\; -->3{)}^2$.
My next step was to equate my derivative to zero, but that does not seem to work as my x cancels out. Usually I would take the x-value(worked out by equating the derivative with zero) and substitute it into the original equation to get a y-value. This would then be the critical points. Is there anyone who could maybe help me out (maybe with an example or so) as I also have to find the intervals where the function is increasing and decreasing?

Is f(x) increasing or decreasing? on $(-<!--\; -\; -->1,0]$
Let the function f(x) be defined by $f(x)=\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{(-<!--\; -\; -->1{)}^{n-<!--\; -\; -->1}}{(2n+1)!}{x}^n$. Since f(x) is a function, then it can increase and decrease. So, in the interval $(-<!--\; -\; -->1,0]$, is f(x) increasing or decreasing? Here we restrict x to any real number between this interval. Fortunately, I found from a previous question I asked that when $x<0$, $f(x)=\frac{\sinh <!--\; \; -->(\sqrt{x})}{\sqrt{x}}$. Credits to the people that answered that question. Graphing this function, I found that it is decreasing throughout. Of course, at $x=0$, the entire function is 0, and it is neither decreasing or increasing.

Find the interval in which f(x) increases and decreases.
Let $f(x)=2x^3-<!--\; -\; -->9x^2+12x+6$ so $f^{\prime }(x)=6x^2-<!--\; -\; -->18x+12=6(x-<!--\; -\; -->1)(x-<!--\; -\; -->2)$.
I need the intervals in which f(x) strictly increases, $f^{\prime }(x)>0$ when $x<1$ and $x>2$ and thus f(x) strictly increases in these intervals and $f^{\prime }(x)<0$ when $1<x<2$ so f(x) strictly decreases in this interval.
My Question:
What about at points $x=1,2$.
If $f^{\prime }(x)=0$ at points (not intervals) then f(x) can still be considered strictly monotone. And it also seems reasonable (I'll add the reason below) to include the points 1,2 in the intervals of increase (IOI, for short) and decreases (IOD).
Eg: Take $x=1$. Let's say I include this point in both IOI and IOD. So IOI is now, $x\in <!--\; \in \; -->(-<!--\; -\; -->\infty ,1]$ and you can see that it doesn't contradict the definition of "strictly increasing function in interval" either. Take any $p,q\in <!--\; \in \; -->(-<!--\; -\; -->\infty ,1],p>q\Rightarrow <!--\; \Rightarrow \; -->f(p)>f(q)$ similarly my IOD, now would be $x\in <!--\; \in \; -->[1,2]$ (notice I included 2) and it still follows the definition.
As I understand that, definition of monotonicity functions at a point, would now get in the way.
I can't include $x=1$ because there exists no $h>0$ such that taking $p,q\in <!--\; \in \; -->(1-<!--\; -\; -->h,1+h)\nRightarrow <!--\; \nRightarrow \; -->f(p)>f(q)$ similarly, I can't add $x=1$ in interval of decrease either.
If I'd have to pick, my intuition would lead me to pick the second one, but I can't see why I should reject the first one either since it actually follows the definition of strictly increasing function in interval.

Is $f(x)=x^{5/3}-5x^{2/3}$ defined over $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]$?
I've encountered this question: Find and describe all local extrema of $f(x)=x^{5/3}-5x^{2/3}.$.
Also indicate on which regions the function is increasing and decreasing.
I've managed to find the extrema, but I am not sure whether the function is defined on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]$. To make sure I looked on the internet at some graphing calculators and some of them graphed the function on that interval while others did not. Is it defined on that interval?

Why is 2 double root of the derivative?
A polynomial function P(x) with degree 5 increases in the interval $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)$ and $(3,\mathrm{\infty <!--\; \infty \; -->})$ and decreases in the interval (1,3). Given that $P^{\prime }(2)=0$ and $P(0)=4$, find P'(6).
In this problem, I have recognised that 2 is an inflection point and the derivative will be of the form: $P^{\prime }(x)=5(x-<!--\; -\; -->1)(x-<!--\; -\; -->3)(x-<!--\; -\; -->2)(x-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->})$.
But I am unable to understand why $x=2$ is double root of the derivative (i.e. why is $\mathrm{\alpha <!--\; \alpha \; -->}=2$?). It's not making sense to me. I need help with that part.