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

How can one determinate the variationt of f(g(x))
Given the two functions:
$f(x)=x^2-<!--\; -\; -->2x$
$g(x)=\sqrt{x+1}$
The question is determinate the variation of f(g(x))
We have $D_{f(g(x))}=(-<!--\; -\; -->1,+\mathrm{\infty <!--\; \infty \; -->})$
For f it's decreasing for $x<1$.
Increasing for $x>1$
For g its increasing for $x>-<!--\; -\; -->1$
To determinate the variationf of f(g(x))
In interval $(-<!--\; -\; -->1;+\mathrm{\infty <!--\; \infty \; -->})$
We have g is increasing
$X>-<!--\; -\; -->1$ means that $g(x)>g(-<!--\; -\; -->1)$ so $g(x)>0$.
So $g([-<!--\; -\; -->1;+\mathrm{\infty <!--\; \infty \; -->}))=[0,+\mathrm{\infty <!--\; \infty \; -->})$.
But the problem is that f in that interval is increasing and decreasing Im stuck here.
Can one write a methode to answer any question like this theoricaly.

Graphing functions with calculus
I am stuck on graphing $y=(x-<!--\; -\; -->3)\sqrt{x}$
I am pretty sure that the domain is all positive numbers including zero. The y intercept is 0, x is 0 and 3.
There are no asymptotes or symmetry.
Finding the interval of increase or decrease I take the derivative which will give me $\sqrt{x}+\frac{x-<!--\; -\; -->3}{2\sqrt{x}}$.
Finding zeroes for this I subtract $\sqrt{x}$ and then multiply by the denominator $2x=x-<!--\; -\; -->3$.
This gives me 3 as a zero, but this isn't correct according to the book so I am stuck. I am not sure what is wrong.

Determine convergence / divergence of $\sum <!--\; \sum \; -->\sin <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{n^2}$.
Let $a_n=\sin <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{n^2}$.
I attempted the integral test but on the interval $[1,\sqrt{2})$ it is increasing and decreasing on $(\sqrt{2},\mathrm{\infty <!--\; \infty \; -->})$. So the integral test in only applicable for the decreasing part. + the integral computation seems to lead to 3 pages of steps...
I believe the comparison test would be the most reasonable test, graphically I observed that $a_n$ behaves like $b_n=1/n$ when n is large.
$\sum <!--\; \sum \; -->b_n=\mathrm{\infty <!--\; \infty \; -->}$; but, since $b_n>a_n⟹<!--\; ⟹\; -->$ inconclusive for divergence/convergence.
I attempted the limit comparison, and ratio test, but inconclusive. I am uncertain if I am doing them properly.
How could I bound below $a_n$ to proceed with the comparison test? Is there a more appropriate method? How would you proceed?

Find the percent of increase.
Original price:$4
New price:$4.50

Finding f intervals
How would I find the f intervals for the following two functions.
$f(x)=(x-<!--\; -\; -->2{)}^2(x+1{)}^2$
using the chain rule I got $(x-<!--\; -\; -->2{)}^2(2)(x+1)(1)+(2)(x-<!--\; -\; -->2)(1)(x+1{)}^2$
then I got f decrease $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->1]$
and f increase $[2,\mathrm{\infty <!--\; \infty \; -->})$
but the area between -1 and 2 in confusing me.
My second function is $f(x)=x+\frac{4}{x^2}$
differentiating I got $\frac{x^3-<!--\; -\; -->8}{x^3}$
so I got f increase $(\mathrm{\infty <!--\; \infty \; -->},0)$ and $[2,\mathrm{\infty <!--\; \infty \; -->})$
f decrease (0, 2]
but did I do this correctly.

Proving the existence of a non-monotone continuous function defined on [0,1]
Let $(I_n{)}_{n\in <!--\; \in \; -->\mathbb{N}}$ be the sequence of intervals of [0,1] with rational endpoints, and for every $n\in <!--\; \in \; -->\mathbb{N}$ let $E_n=\{f\in <!--\; \in \; -->C[0,1]:f\text{is monotone in}{I}_n\}$. Prove that for every $n\in <!--\; \in \; -->\mathbb{N}$, $E_n$ is closed and nowhere dense in $(C[0,1],d_{\mathrm{\infty <!--\; \infty \; -->}})$. Deduce that there are continuous functions in the interval [0,1] which aren't monotone in any subinterval.
For a given n, $E_n$ can be expressed as $E_n=E_{n\nearrow <!--\; \nearrow \; -->}\cup <!--\; \cup \; -->E_{n\swarrow <!--\; \swarrow \; -->}$ where $E_{n\nearrow <!--\; \nearrow \; -->}$ and $E_{n\swarrow <!--\; \swarrow \; -->}$ are the sets of monotonically increasing functions and monotonically decreasing functions in $E_n$ respectively. I am having problems trying to prove that these two sets are closed. I mean, take $f\in <!--\; \in \; -->(C[0,1],d_{\mathrm{\infty <!--\; \infty \; -->}})$ such that there is $\{f_k{\}}_{k\in <!--\; \in \; -->\mathbb{N}}\subset <!--\; \subset \; -->E_{n\nearrow <!--\; \nearrow \; -->}$ with $f_k\to <!--\; \to \; -->f$. How can I prove $f\in <!--\; \in \; -->E_{n\nearrow <!--\; \nearrow \; -->}$?. Suppose I could prove this, then I have to show that ${\stackrel{¯<!--\; ¯\; -->}{E_n}}^{\circ <!--\; \circ \; -->}=E_n^{\circ <!--\; \circ \; -->}=\mathrm{\varnothing <!--\; \varnothing \; -->}$. This means that for every $f\in <!--\; \in \; -->E_n$ and every $r>0$, there is $g\in <!--\; \in \; -->B(f,r)$ such that g is not monotone. Again, I am stuck. If I could solve this two points, it's not difficult to check the hypothesis and apply the Baire category theorem to prove the last statement.

Intermediate value theorem on infinite interval $\mathbb{R}$
I have a continious function f that is strictly increasing. And a continious function g that is strictly decreasing. How to I rigorously prove that $f(x)=g(x)$ has a unique solution?
Intuitively, I understand that if I take limits to infinity, then f grows really large and g grows very small so the difference is less than zero. If I take the limits to negative infinity then the opposite happens. Using the intermediate value theorem, there must be an intersection, and since they are strictly increasing/decreasing, only one intersection happens.
My question is how do I apply the intermediate value theorem here? I don't have the interval to apply it on. I don't know when f crosses g and therefore can't take any interval. Or is there some sort of axiom applied here that I am missing?

Sequence defined by $u_0=1/2$ and the recurrence relation $u_{n+1}=1-<!--\; -\; -->u_n^2$.
I want to study the sequence defined by $u_0=1/2$ and the recurrence relation $u_{n+1}=1-<!--\; -\; -->u_n^{2}n\ge <!--\; \ge \; -->0.$.
I calculated sufficient terms to understand that this sequence does not converge because its odd and even subsequences converge to different limits.
In particular we have $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{a}_{2n+1}=0\text{ and}\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{a}_{2n+2}=1.$.
Now I can also prove that $u_n\le <!--\; \le \; -->1$ because the subsequent terms are related in this way
$f:<!--\; :\; -->x\mapsto <!--\; \mapsto \; -->1-<!--\; -\; -->x^2.$
In order to prove that the subsequences converge I need to prove either that they are bounded (in order to use Bolzano-Weierstrass) or that they are bounded and monotonic for $n>N$. Proved that the subsequences converge to two different limits, I will be able to prove that our $u_n$ diverges.
Can you help me? Have you any other idea to study this sequence?

Derivative of an integral from 0 to x when x is negative?
Given a function $F(x)={\int <!--\; \int \; -->}_0^x\frac{t+8}{t^3-<!--\; -\; -->9}dt,$, is F′(x) different when $x<0$, when $x=0$ and when $x>0$?
When $x<0$, is $F^{\prime }(x)=-<!--\; -\; -->\frac{x+8}{x^3-<!--\; -\; -->9}$.
... since you can't evaluate an integral going from a smaller number to a bigger number? That's what I initially thought, but when I graphed the antiderivative the intervals of increase and decrease were different from the ones in my calculation.
EDIT: Yes, I was talking about that identity. So is $F^{\prime }(x)=\frac{x+8}{x^3-<!--\; -\; -->9}$ or $F^{\prime }(x)=-<!--\; -\; -->\frac{x+8}{x^3-<!--\; -\; -->9}$.

Find min/max along with concavity, inflection points, and asymptotes
For the function $ƒ(x)=\frac{x^2}{(x-2{)}^2}$
I know the derivative is $f^{\prime }(x)=-<!--\; -\; -->\frac{4x}{{(x-<!--\; -\; -->2)}^3}$ and $f^{″}(x)=\frac{8(x+1)}{{(x-<!--\; -\; -->2)}^4}$.
Critical point is $x=-<!--\; -\; -->1$ which is also the min.
I think possible inflection point are $x=1$ and $x=0$.
I need to find: 1)Find the vertical and horizontal asymptotes of ƒ(x).
2) Find the intervals on which ƒ(x)‍ is increasing or decreasing.
3) Find the critical numbers and specify where local maxima and minima of ƒ(x) occur.
4) Find the intervals of concavity and the inflection points of ƒ(x).

Prove that ${\mathrm{\chi <!--\; \chi \; -->}}_{(0,1)}-<!--\; -\; -->{\mathrm{\chi <!--\; \chi \; -->}}_S$ is not a limit of increasing step functions.
Let $A_n$ be sequence of connected bounded subsets (interval) of real numbers. Step function is defined to be the finite linear combination of their charateristic functions.
$\mathrm{\psi <!--\; \psi \; -->}=c_1{\mathrm{\chi <!--\; \chi \; -->}}_{A_1}+c_2{\mathrm{\chi <!--\; \chi \; -->}}_{A_2}+...+c_n{\mathrm{\chi <!--\; \chi \; -->}}_{A_n}$
while $c_k\in <!--\; \in \; -->\mathbb{R}$ for $k=1,2,...,n$.
Let $I_n$ be a sequence of open intervals in (0,1) which covers all the rational points in (0,1) and such that $\sum <!--\; \sum \; -->\int <!--\; \int \; -->{\mathrm{\chi <!--\; \chi \; -->}}_{I_n}\le <!--\; \le \; -->\frac{1}{2}$. Let $S=\bigcup <!--\; \bigcup \; -->I_n$ and $f={\mathrm{\chi <!--\; \chi \; -->}}_{(0,1)}-<!--\; -\; -->{\mathrm{\chi <!--\; \chi \; -->}}_S$ show that there is no increasng sequence of step function $\{{\mathrm{\psi <!--\; \psi \; -->}}_n\}$ such that $lim{\mathrm{\psi <!--\; \psi \; -->}}_n(x)=f(x)$ almost everywhere. (by means of increasing, ${\mathrm{\psi <!--\; \psi \; -->}}_n(x)\le <!--\; \le \; -->{\mathrm{\psi <!--\; \psi \; -->}}_{n+1}(x)$ for all x)
I think ${\mathrm{\psi <!--\; \psi \; -->}}_n={\mathrm{\chi <!--\; \chi \; -->}}_{(0,1)}-<!--\; -\; -->\underset{k=1}{\overset{n}{\sum <!--\; \sum \; -->}}{\mathrm{\chi <!--\; \chi \; -->}}_{I_k}$ is what author intended. $\int <!--\; \int \; -->{\mathrm{\psi <!--\; \psi \; -->}}_n$ is decreasing and $\int <!--\; \int \; -->{\mathrm{\psi <!--\; \psi \; -->}}_n\ge <!--\; \ge \; -->1-<!--\; -\; -->\frac{1}{2}$. this shows that $lim\int <!--\; \int \; -->{\mathrm{\psi <!--\; \psi \; -->}}_n$ converges. so $lim{\mathrm{\psi <!--\; \psi \; -->}}_n(x)=f(x)$ almost everywhere. However convergence of ${\mathrm{\psi <!--\; \psi \; -->}}_n$ doesn't prove the non-existence of increasing step function which converges to f a.e.
How can I finish the proof?

How to find where a function is increasing at the greatest rate
Given the function $f(x)=\frac{1000x^2}{11+x^2}$ on the interval [0,3], how would I calculate where the function is increasing at the greatest rate?

Is a horizontal line an increasing or decreasing function?
This is the definition of an increasing and decreasing function.
"A function f(x) increases on an interval I if $f(b)\ge f(a)\mathrm{\forall <!--\; \forall \; -->}b>a$, where $a,b\in <!--\; \in \; -->I$. If $f(b)>f(a)\mathrm{\forall <!--\; \forall \; -->}b>a$, the function is said to be strictly increasing.
Conversely, A function f(x) decreases on an interval I if $f(b)\le f(a)\mathrm{\forall <!--\; \forall \; -->}b>a$, where $a,b\in <!--\; \in \; -->I$. If $f(b)<f(a)\mathrm{\forall <!--\; \forall \; -->}b>a$, the function is said to be strictly decreasing.
Then how would a horizontal line be described? If $f(x_2)=f(x_1)\mathrm{\forall <!--\; \forall \; -->}{x}_2>x_1$ does that mean that a horizontal line meets the definition of an increasing function and a decreasing function?

Stability, critical points and similar properties of solutions of nonlinear Volterra integral equations
I have a system of nonlinear Volterra integral equations of form $x(t)=x_0+{\int <!--\; \int \; -->}_0^{t}K(t,s)F(x(s))ds$ and I am interested on the critical points of x(t), I mean maximum, minimum, increasing and decreasing intervals, nonnegativity etc.
I imagine it's impossible to get complete informations about that, but here I am asking for theorems and general results to help me to study these aspects, once is impossible know the true solution.

Concavity and critical numbers
I am attempting to study for a test, but I forgot how to do all the stuff from earlier in the chapter.
I am attempting to find the intervals where f is increasing and decreasing, min and max values and intervals of concavity and inflection points.
I have $f(x)=e^{2x}+e^{-<!--\; -\; -->x}$. I know that I have to find the derivative, set to zero to find the critical numbers then test points then find the second derivative and do the same to find the concavity but I don't know how to get the derivative of this. Wolfram Alpha gives me a completely different number than what I get.

Question on Local Maxima and Local Minima
Find the set of all the possible values of a for which the function $f(x)=5+(a-<!--\; -\; -->2)x+(a-<!--\; -\; -->1)x^2-<!--\; -\; -->x^3$ has a local minimum value at some $x<1$ and local maximum value at some $x>1$.
The first derivative of f(x) is:
$f^{\prime }(x)=(a-<!--\; -\; -->2)+2x(a-<!--\; -\; -->1)-<!--\; -\; -->3x^2$
I do know the first derivative test for local maxima and local minima, but I can't figure out how I could use monotonicity to find intervals of increase and decrease of f′(x)
The expression for f′(x) might suggest the double derivative test is the key, considering $f^{″}(x)=2(a-<!--\; -\; -->1)-<!--\; -\; -->6x$ for which the intervals where it is greater than zero and less than zero can be easily found, but then again I can't think of a way how I could find a c such that $f^{\prime }(c)=0$.

First Derivative vs Second Derivative in relation to minima and maxima
I am struggling with understanding the difference between them and I need to write about them intuitively. The way my teacher explained it, the sign of the first derivative is used to determine if there is a minimum, maximum, or neither. For example, if the derivative is increasing to the left of x and decreasing to the right of x, than a maximum is present. If the derivative is decreasing to the left of x and is increasing to the right of x, there is a minimum. I know that the second derivative is the derivative if the first derivative and if it is positive on a certain interval, it is concave up. Is that not telling me the same thing as the first derivative?

If $1\le <!--\; \le \; -->\mathrm{\alpha <!--\; \alpha \; -->}$ show show that the gamma density has a maximum at $\frac{\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->1}{\mathrm{\lambda <!--\; \lambda \; -->}}$.
So using this form of the gamma density function:
$g(t)=\frac{{\mathrm{\lambda <!--\; \lambda \; -->}}^{\mathrm{\alpha <!--\; \alpha \; -->}}{t}^{\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->1}{e}^{-<!--\; -\; -->\mathrm{\lambda <!--\; \lambda \; -->}t}}{{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}{t}^{\mathrm{\alpha <!--\; \alpha \; -->}-<!--\; -\; -->1}{e}^{-<!--\; -\; -->t}dt}$
I would like to maximize this. Now i was thinking of trying to differentiate with respect to t, but that is becoming an issue because if i remember correctly the denominator is not a close form when the bound goes to infiniti

Sum of monotonic increasing and monotonic decreasing functions
Consider an interval $x\in <!--\; \in \; -->[x_0,x_1]$. Assume there are two functions f(x) and g(x) with $f^{\prime }(x)\ge <!--\; \ge \; -->0$ and $g^{\prime }(x)\le <!--\; \le \; -->0$. We know that $f(x_0)\le <!--\; \le \; -->0$, $f(x_1)\ge <!--\; \ge \; -->0$, but $g(x)\ge <!--\; \ge \; -->0$ for all $x\in <!--\; \in \; -->[x_0,x_1]$. I want to show that $q(x)\equiv <!--\; \equiv \; -->f(x)+g(x)$ will cross zero only once. We know that $q(x_0)\le <!--\; \le \; -->0$ and $q(x_1)\ge <!--\; \ge \; -->0$.

Find integral values k such that sum of expression is minimized
Given n values $X_1,X_2,....,X_n$, where $X_i$ can be positive or negative.
The absolute values of $X_i$ will be less than 100000, also $n<=100000$.
What should be the possible value(s) of k such that $f(k)=|k|+|X_1+k|+|X_1+X_2+k|+\cdots <!--\; \cdots \; -->+|X_1+X_2+X_3+\cdots <!--\; \cdots \; -->+X_n+k|$ is minimized .
$|a|$ absolute value of a .