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

Arrange the following in the correct order of increasing density.
Air
Oil
Water
Brick

How to evaluate the integral ${\displaystyle \int \frac{dx}{x^3+1}}$?

Let h be a twice differentiable positive function on an open interval J. Let g(x)=ln(h(x)) for each x∈J. Suppose $(h\prime (x){)}^2>h\prime \prime (x)h(x)$ for each x∈J. Then
A) g is increasing on J
B) g is decreasing on J
C) g is concave up on J
D) g is concave down on J

Interval Notation for Increasing and Decreasing Intervals of a Function
This was brought up by another student in one of my pre-calculus classes.
The graph was a simple quadratic $x^2$. The teacher stated that the graph was decreasing from $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0)$, and increasing from $(0,\mathrm{\infty <!--\; \infty \; -->})$.
Why would zero not be included? i.e: decr. $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]$ and incr. $[0,\mathrm{\infty <!--\; \infty \; -->})$

Find increasing and decreasing intervals of a function
The given function is $f(x)=x^{200}-<!--\; -\; -->x^{100}$, and I'm supposed to find it's decreasing and increasing intervals. Also, I should find them not by using derivatives but by doing function composition, like this:
$f_1(x)=x^{100}$
$f_2(x)=x(x-<!--\; -\; -->1)$
$f(x)=f_2(f_1(x))$
I know that $f_1$ is decreasing on the interval $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},0]$ and increasing on $[0,+\mathrm{\infty <!--\; \infty \; -->})$, and $f_2$ is decreasing on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\frac{1}{2}]$ and increasing on $[\frac{1}{2},+\mathrm{\infty <!--\; \infty \; -->})$, but I'm not really sure what to do next.

Negation of monotonicity of a continuous function
So I am given a continuous function mapping a connected domain to the reals, i.e $f:(a,b)\to <!--\; \to \; -->\mathbb{R}$.
I want to show that if f is not strictly monotone and f is continuous, we have $x,y,z\in <!--\; \in \; -->(a,b)$ with $x<y<z$ such that:
$f(x)\le <!--\; \le \; -->f(y)\text{ and }f(y)\ge <!--\; \ge \; -->f(z)$
or $f(x)\ge <!--\; \ge \; -->f(y)\text{ and }f(y)\le <!--\; \le \; -->f(z)$
As for what I've tried, using the negation of strictly monotone, we have that f must be increasing and decreasing in two intervals $[x_1,x_2]$ and $[x_3,x_4]$ i.e.:
$f(x_1)\le <!--\; \le \; -->f(x_2)\text{ and }f(x_3)\ge <!--\; \ge \; -->f(x_4)$ or $f(x_1)\ge <!--\; \ge \; -->f(x_2)\text{ and }f(x_3)\le <!--\; \le \; -->f(x_4)$
The problem comes when I look to combine the two into a single inequality. In an effort to combine these, the only way I see of doing it (directly) would be casewise. If anyone could provide a hint or (less preferred) a complete direct proof I would appreciate it.

Where is this function concave up?
The function is: $(x^2-<!--\; -\; -->16{)}^6$.
So I did all the relevant calculations and all my others answers about inflexion points, critical points, extreme points, concave down, increasing and decreasing intervals were correct. But my interval for concave up keeps coming up as wrong.
This is the interval I calculated:
$(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->4)\cup <!--\; \cup \; -->(-<!--\; -\; -->4,-<!--\; -\; -->\frac{4\sqrt{11}}{11})\cup <!--\; \cup \; -->(\frac{4\sqrt{11}}{11},4)\cup <!--\; \cup \; -->(4,\mathrm{\infty <!--\; \infty \; -->})$
I also tried various alternative ones like:
$(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\frac{4\sqrt{11}}{11})\cup <!--\; \cup \; -->(\frac{4\sqrt{11}}{11},\mathrm{\infty <!--\; \infty \; -->})$

How can I prove the monotonicity of a function?
Given is the function $f(x)=\frac{x}{\sqrt{1-<!--\; -\; -->x^2}}$.
How can I prove that this function is monotonic and thus injective?

Finding the counterclockwise nearest point in $\mathcal{O}(\log <!--\; \; -->(n))$ time
Given a point q, direction w and point set P with convex hull $(v_1,\dots <!--\; \dots \; -->,v_n)$ known, give an algorithm to find the point in P that (q,w) touches first when rotated counterclockwise in $\mathcal{O}(\log <!--\; \; -->(n))$ time.
REMARK: We can assume that $(v_1,\dots <!--\; \dots \; -->,v_n)$ is sorted counterclockwise.
My idea would be the following: Since $(v_1,\dots <!--\; \dots \; -->,v_n)$ is sorted we can modify the approach of Binary Search to get the required $\mathcal{O}(\log <!--\; \; -->(n))$ bound. I suppose we have to test the angles between $(q,w,v_i)$ repeatedly, but I do not see how to formulate this clearly.

Graph of $\frac{e^x-<!--\; -\; -->1}{x}$
i am trying to draw the graph of $f(x)=\frac{e^x-<!--\; -\; -->1}{x}$.
its domain is $\mathbb{R}$-0 but its limit is 1 in the neighbourhood of zero.
Also $li{m}_{x\to <!--\; \to \; -->-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}}f(x)=0$ so negative X axis is asymptote.
also $\frac{df(x)}{dx}=\frac{(x-<!--\; -\; -->1)e^x+1}{x^2}$
so f(x) is increasing in $(0\mathrm{\infty <!--\; \infty \; -->})$ and decreasing in $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}0)$
now the ambiguity is $f(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->})\to <!--\; \to \; -->0$ and $f(0^{-<!--\; -\; -->})\to <!--\; \to \; -->1$ but f(x) is decreasing in negative interval.

Question about finding where the function increases and decreases on $f(x)=\frac{1}{x}$.
$f(x)=\frac{1}{x},x\ge <!--\; \ge \; -->1$
I have been staring at this equation for a bit. Things I'm confused on.
the derivative of this is: $f^{\prime }(x)=\frac{-<!--\; -\; -->1}{x^2}$ now, how am I supposed to find where this derivative increases/decreases? Do I find the critical points first? by setting the derivative to 0? or do I solve it like $\frac{-<!--\; -\; -->1}{x^2}>0$ cross multiply to make it: $-<!--\; -\; -->1>x^2$ and if so once I square this does it make the result $x=-<!--\; -\; -->1,x=1$? I'm really lost here and it seems like it should be easier.
Does setting the derivative to > or < or = and solving for the x give a critical point?

Calculus application of derivatives
Let $g(x)=2f(x/2)+f(2-<!--\; -\; -->x)$ and $f^{″}(x)<0$ for all $x\in <!--\; \in \; -->(0,2)$. If g(x) increases in (a,b) and decreases in (c,d) find the values of a, b, c and d.
What I thought was a little graphical approach. I figured it out that if f′′(x) is less than zero then f′(x) would be decreasing (not considering the concavity) and the behaviour of g(x) would depend upon f(x). But I couldn't figure out the exact intervals.

Finding the intervals of increase and decrease of $\frac{x^4-<!--\; -\; -->x^3-<!--\; -\; -->8}{x^2-<!--\; -\; -->x-<!--\; -\; -->6}$.
I tried to find the derivative by the quotient rule to obtain the critical points but the formula was getting complicated, I know that $D_f=\mathbb{R}\setminus <!--\; \setminus \; -->\{-<!--\; -\; -->2,3\}$ but then what?

How do I get an integer from a polygon equation, which usually returns fractions?
Known: edge length
Unknown: number of edges
Radius should increase of decrease to an interval to ensure number of edges in Polygon is divisible by 1
E.g. edge length is 100mm, number of edges unknown, radius is 6m. How do I find the closest radius to 6m which gives an integer, divisible by 1, for a realistic number of edges.
Previous Question
If a regular polygon has a fixed edge length, can I know how many edges it has by knowing the length from corner to its center?
and it's answer
the radius of the polygon, and it has the formula
$r=\frac{s}{2\sin <!--\; \; -->\left(\frac{180°}{n}\right)}$
where s is the side length of the polygon and n is the number of sides. So given r and s, you can simply solve the above equation for n.
It's worth pointing out that when you solve for n there's no guarantee that it will turn out to be an integer, and hence correspond to a regular polygon
I have no math background, not even enough to know which tags to attach. How do I solve for these intervals of radii?

Interval of monotonicity of the function $f(x)=\frac{\ln <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}+x)}{\ln <!--\; \; -->(e+x)}$.
Find the exact interval in which the function $f(x)=\frac{\ln <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}+x)}{\ln <!--\; \; -->(e+x)}$ is increasing and decreasing.
My try: The domain of the function is $(-<!--\; -\; -->e,\mathrm{\infty <!--\; \infty \; -->})$
The derivative is: $f^{\prime }(x)=\frac{(e+x)\ln <!--\; \; -->(e+x)-<!--\; -\; -->(\mathrm{\pi <!--\; \pi \; -->}+x)\ln <!--\; \; -->(\mathrm{\pi <!--\; \pi \; -->}+x)}{(\mathrm{\pi <!--\; \pi \; -->}+x)(e+x){\ln }^2<!--\; \; -->(e+x)}$.
Since due to the domain we have $e+x>0,\mathrm{\pi <!--\; \pi \; -->}+x>0$ and trivially ${\ln }^2<!--\; \; -->(e+x)>0$.
Now it is sufficient to check the nature of numerator above.
Let $p=e+x,q=\mathrm{\pi <!--\; \pi \; -->}+x$.
The numerator is in the form $p\ln <!--\; \; -->(p)-<!--\; -\; -->q\ln <!--\; \; -->(q)$ where $p<q,\mathrm{\forall <!--\; \forall \; -->}x$.
Now we know that the function $h(x)=x\ln <!--\; \; -->x$ is increasing in $[\frac{1}{e},\mathrm{\infty <!--\; \infty \; -->})$.
Case 1. If $\frac{1}{e}\le <!--\; \le \; -->p<q$ we have:
$h(p)<h(q)⟹<!--\; ⟹\; -->p\ln <!--\; \; -->p-<!--\; -\; -->q\ln <!--\; \; -->q<0$
Thus $f^{\prime }(x)<0,\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->[\frac{1}{e}-<!--\; -\; -->e,\mathrm{\infty <!--\; \infty \; -->})=[-<!--\; -\; -->2.35,\mathrm{\infty <!--\; \infty \; -->})$
So f(x) must be decreasing in the above interval. But i compared f(-2) and f(-1) It is the other way round, that is $f(-<!--\; -\; -->2)<f(-<!--\; -\; -->1)$
Now what i did is i used graphing calculator to check the graph. Yes it is decreasing, but there is a vertical asymptote. How to identify the equation of vertical asymptote?
Also the graph is decreasing in $[-<!--\; -\; -->2.541,\mathrm{\infty <!--\; \infty \; -->})$ which is not matching with my answer above.

First derivative test to find where the function is increasing, and decreasing
I have this function: $f(x)=\frac{x^2}{log(x)-<!--\; -\; -->1}$.
and I want to find increasing intervals, and decreasing intervals. Here's what I did:
- I've found domain in which the function is defined, and this is what I got: $f:]0,+\mathrm{\infty <!--\; \infty \; -->}[(without1)-<!--\; -\; -->>R$.
0 and 1 are critical points in which the function isn't defined (i.e if $x=0$, the log is undefined, and if $x=1$, the fraction is undefined).
- I've computed first derivative, and this is what I got (I've used quotient rule):
$\frac{2x\ast <!--\; \ast \; -->(log(x)-<!--\; -\; -->1)-<!--\; -\; -->x^2\ast <!--\; \ast \; -->1/x}{(log(x)-1{)}^2}$
I've rewritten it, as follows: $\frac{2x\ast <!--\; \ast \; -->(log(x)-<!--\; -\; -->1)-<!--\; -\; -->x}{(log(x)-1{)}^2}$.
the denominator is always positive so the sign depends only on the numerator. (I think the problem is in this step) on a line, I've chosen 4 different random numbers, 1/4, 1/2, 3/2, 2, and I've put each number in the first derivative, and this is what I got:
- in the first interval (from 0, to 1/4), the original function is decreasing (minus sign)
- same for the other intervals
I don't have the full solution of the exercise, therefore I don't knwo if my solution is correct, but I think is strange that in this function there aren't increasing intervals.
- I think the problem is in the first derivative, but I'm sure I've used the quotient rule correctly.

Find the interval of increase and decrease ,local extremes , inflection points of $f(x)=x^{\frac{4}{3}}-<!--\; -\; -->x^{\frac{1}{3}}$
Since $f(x)=x^{\frac{4}{3}}-<!--\; -\; -->x^{\frac{1}{3}}\Rightarrow <!--\; \Rightarrow \; -->f^{\prime }(x)=\frac{4}{3}{x}^{\frac{1}{3}}-<!--\; -\; -->\frac{1}{3}{x}^{\frac{-<!--\; -\; -->1}{3}}$

Intervals in which f(x) is Strictly Increasing/Decreasing
Find the intervals in which $f(x)=\sin <!--\; \; -->x+\cos <!--\; \; -->x,0\le <!--\; \le \; -->x\le <!--\; \le \; -->2\mathrm{\pi <!--\; \pi \; -->}$ is strictly increasing/decreasing.
First I find the derivative $f^{\prime }(x)=\cos <!--\; \; -->x-<!--\; -\; -->\sin <!--\; \; -->x$, then put $f^{\prime }(x)=0$, getting $\tan <!--\; \; -->x=1$. The principal solutions of $\tan <!--\; \; -->x=1$ are $x=\mathrm{\pi <!--\; \pi \; -->}/4$ and $x=5\mathrm{\pi <!--\; \pi \; -->}/4$, which gives the intervals $[0,\mathrm{\pi <!--\; \pi \; -->}/4)$, $(\mathrm{\pi <!--\; \pi \; -->}/4,5\mathrm{\pi <!--\; \pi \; -->}/4)$, and $(5\mathrm{\pi <!--\; \pi \; -->}/4,2\mathrm{\pi <!--\; \pi \; -->})$.
After this I am stuck. The book just solves the question by making a table showing the interval, sign of the derivative as positive or negative, and strictly increasing for positive and vice versa.
I just want to know how do they get it positive or negative.It would really help if someone did one for the interval $(\mathrm{\pi <!--\; \pi \; -->}/4,5\mathrm{\pi <!--\; \pi \; -->}/4)$.

Intervals on which function is increasing and decreasing
Let $p(x)=x^5-<!--\; -\; -->q^{2}x-<!--\; -\; -->q$, where q is a prime number. I want to understand how to determine when the function will be decreasing and increasing on the intervals given below.
We compute $p^{\mathrm{\prime <!--\; \prime \; -->}}(x)=5x^4-<!--\; -\; -->q^2$ and look for the critical points.
$5x^4-<!--\; -\; -->q^2=0⟺<!--\; ⟺\; -->x=\pm <!--\; \pm \; -->\frac{\sqrt{q}}{\sqrt[4]{5}}$
Hence we have to investigate the behavior of p′(x) for each of these intervals $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->\frac{\sqrt{q}}{\sqrt[4]{5}})$, $(-<!--\; -\; -->\frac{\sqrt{q}}{\sqrt[4]{5}},\frac{\sqrt{q}}{\sqrt[4]{5}})$ and $(\frac{\sqrt{q}}{\sqrt[4]{5}},\mathrm{\infty <!--\; \infty \; -->})$ this will indicate when the function will be increasing and decreasing. How can this be determined when the expression $\frac{\sqrt{q}}{\sqrt[4]{5}}$ contains a prime number???
The answer should be : the function will be increasing for $x<\frac{\sqrt{q}}{\sqrt[4]{5}}$ and strictly decreasing for $-<!--\; -\; -->\frac{\sqrt{q}}{\sqrt[4]{5}}<x<\frac{\sqrt{q}}{\sqrt[4]{5}}$ and strictly increasing again for $x>\frac{\sqrt{q}}{\sqrt[4]{5}}$.

Prove the integral inequality
$\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}-<!--\; -\; -->\frac{1}{2}\le <!--\; \le \; -->{\int <!--\; \int \; -->}_0^1\frac{\arcsin <!--\; \; -->x}{1+x^8}\le <!--\; \le \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->1$
$\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}-<!--\; -\; -->\frac{1}{2}\le <!--\; \le \; -->{\int <!--\; \int \; -->}_0^1\frac{\arcsin <!--\; \; -->x}{1+x^8}\le <!--\; \le \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->1$
I can say that I have close to zero experience with integral inequalities. I've searched for couple of theorems most of them involving monotonous function on [0,1]. Arcsin is increasing in this case and the other one is decreasing. You could say we have a product of increasing and decreasing function on this interval however I didn't find any theorems that cover this case. Also this integral doesn't seem to have elementary function.
Another thing I noticed is that most of the theorems cover only 1 part of this inequality; either $\le <!--\; \le \; -->or\ge <!--\; \ge \; -->$ so my guess is we need some type of combination or perhaps find a new integral that is equal to $\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}-<!--\; -\; -->\frac{1}{2}$ and another that is equal to $\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}-<!--\; -\; -->1$.
What's the correct approach when we need to prove that the value of certain integral is in a provided interval?