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

What would the critical points be of this equation?
${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{-<!--\; -\; -->4x}{(1+x^2{)}^2}}$
I just dont understand how find them with a reciprocal function. If someone could explain how to find it in this this situation that would be fantasic!

If a rock is thrown upward o the planet Mars with a velocity of 10m/s, its height in meters t seconds later is given by $y=10t-<!--\; -\; -->1.86t^2$
a) Find the average velocity over the given time intervals:
i)[1,2] ii)[1,1.5] iii)[1,1.1] iv)[1,1.01] v) [1,1.001]
b) Estimate the instantaneous velocity when t=1.

Winding number is locally-constant for general curves (not $C^1$) using variation of argument definition
I've been searching for this proof, here and on the web too, but it seems like the answer is harder to find than expected. There are many similar questions about this but they all (implicitly or explicitly) suppose continuous differentiability.
First, let me state the definition of winding number:
Let $\mathrm{\gamma <!--\; \gamma \; -->}:[0,1]\to <!--\; \to \; -->{\mathbb{C}}^{\mathbb{\ast <!--\; \ast \; -->}}$ be a continuos function. We know that for such a path, there exists a continuous argument function, that is a function $\mathrm{\varphi <!--\; \varphi \; -->}:[0,1]\to <!--\; \to \; -->\mathbb{R}$ such that $\mathrm{\gamma <!--\; \gamma \; -->}(t)=|\mathrm{\gamma <!--\; \gamma \; -->}(t)|e^{i\mathrm{\varphi <!--\; \varphi \; -->}(t)}$, for all $t\in <!--\; \in \; -->[0,1]$. We define the variation of argument along $\mathrm{\gamma <!--\; \gamma \; -->}$:
$\mathrm{v}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{r}\mathrm{g}(\mathrm{\gamma <!--\; \gamma \; -->})=\mathrm{\varphi <!--\; \varphi \; -->}(1)-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}(0)$
(We can also prove that all such argument functions are of the form $\mathrm{\varphi <!--\; \varphi \; -->}+2n\mathrm{\pi <!--\; \pi \; -->}$, for any integer n, so the variation of argument is well-defined. Also, if $\mathrm{\gamma <!--\; \gamma \; -->}$ is closed, the variation of argument is a multiple of $2\mathrm{\pi <!--\; \pi \; -->}$)
The winding number (index) of the closed curve $\mathrm{\gamma <!--\; \gamma \; -->}$ around 0 is defined as:
$\mathrm{I}(\mathrm{\gamma <!--\; \gamma \; -->},0)=\frac{\mathrm{v}\mathrm{a}\mathrm{r}\mathrm{a}\mathrm{r}\mathrm{g}(\mathrm{\gamma <!--\; \gamma \; -->})}{2\mathrm{\pi <!--\; \pi \; -->}}$
For a closed path $\mathrm{\gamma <!--\; \gamma \; -->}:[0,1]\to <!--\; \to \; -->\mathbb{C}\setminus <!--\; \setminus \; -->\{z_0\}$ for some $z_0\in <!--\; \in \; -->\mathbb{C}$, the winding number of $\mathrm{\gamma <!--\; \gamma \; -->}$ around $z_0$ is defined by translation as:
$\mathrm{I}(\mathrm{\gamma <!--\; \gamma \; -->},z_0)=\mathrm{I}(\mathrm{\gamma <!--\; \gamma \; -->}-<!--\; -\; -->z_0,0)$
I consider this as the best definition of winding number because it is intuitive, general and everything follows from it.
Now I want to prove that the winding number is constant on the connected components of the complement of the curve. This is not very hard to prove using the integral expression of the winding number, but that assumes that $\mathrm{\gamma <!--\; \gamma \; -->}$ is $C^1$.
I know that I should somehow prove that $\mathrm{\varphi <!--\; \varphi \; -->}$ is continuous or just that it is constant in an open ball around any point included in the connected component.
Intuitively, I can prove it like this: consider a curve γ that doesn't go through 0, a point $z_0\ne <!--\; \ne \; -->0$ inside a ball around 0 that is contained in the connected component of 0. Imagine two segments from 0 to the curve and from $z_0$ to the curve, as we move along it. Both segments follow the same point on the circle around 0 and their movements along little circles around their respective start points have the same intervals of increase/decrease in angle (going counter-/clockwise). Therefore, whenever one completes a circle in one direction, the other does so too. Hence, the variation of the argument of $\mathrm{\gamma <!--\; \gamma \; -->}$ around 0 and around $z_0$ is the same. By translation, we generalize the result for other points than 0.
This argument is hard to write rigorously and gets a little too geometric, I think. I wonder if an easier take on this can be carried out.

Absolute value f interval?
How would I find where the interval in which f increases and decreases in the following function.
$f(x)=\mid <!--\; \mid \; -->4-<!--\; -\; -->x^2\mid <!--\; \mid \; -->$
What would I do find the zero
I know $f(x)=\{\begin{array}{r}4-<!--\; -\; -->x^2\text{for}|x|<2\\ x^2-<!--\; -\; -->4\text{for}|x|>2\end{array}$
I suspect there are zeroes at 2 and -2 looking at the graph of the function. But how would find the zeroes.So that I can find the f interval.

Even or Odd symmetry
What type of symmetry does the function $y=\frac{1}{|x|}$ have? Specify the intervals over which the function is increasing and the intervals where it is decreasing.

Conjecture: If $a_k-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(a_k)=0$ then $a_{k+1}/a_k\to <!--\; \to \; -->e$.
A natural question to ask is for what n does $\mathrm{\pi <!--\; \pi \; -->}(n)\mid <!--\; \mid \; -->n$ where $\mathrm{\pi <!--\; \pi \; -->}(n)$ is the prime counting function. It can be shown with a discrete variant of the intermediate value theorem that such an n exists for any $k=n/\mathrm{\pi <!--\; \pi \; -->}(n)$. (Intuition: $f(n)=n-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(k)$ can only increase by 1 and decrease by $k-<!--\; -\; -->1$. Since $f(2)<0$ for $k>3$ and $f(n)>0$ for large n, there must be a zero).
Computations suggest that all solutions to $n-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(n)=0$ are grouped at some $n_k$ in an interval of around $\sqrt{n_k}$ width. But more surprisingly to me, for any pair of solutions $a_k$ and $a_{k+1}$, with the former being in the k group and the latter in the $k+1$ group, the ratio $a_{k+1}/a_k$ seems to be about e.
My conjecture: If $a_k-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(a_k)=0$ then $a_{k+1}/a_k\to <!--\; \to \; -->e$. In other words, the solutions of $f(n)=n-<!--\; -\; -->k\mathrm{\pi <!--\; \pi \; -->}(k)$ grow exponentially as $e^k$.
I am at a loss why this is the case. My guess is the gaps between primes $[n!,n!+n]$ comes into play, giving us Stiring's formula and hence our e factor, but I really have no clue.
Can someone either (1) provide a proof or disproof of this conjecture or (2) give an extremely plausible method of attack? I would also appreciate suggestions in the comments for ideas/theorems/techniques to investigate to help me make my own crack at the problem.

Some doubts with the sign of a derivative
Good evening to everyone. The derivative is defined in the following order:
$\frac{d}{dx}f(x)=\frac{-<!--\; -\; -->x^2-<!--\; -\; -->x+11}{{(x+3)}^2}{e}^{2-<!--\; -\; -->x}$ for $x<-<!--\; -\; -->3$
$\frac{d}{dx}f(x)=\frac{x^2+x-<!--\; -\; -->11}{{(x+3)}^2}{e}^{2-<!--\; -\; -->x}$ for $-<!--\; -\; -->3<x<2$ and $\frac{d}{dx}f(x)=\frac{x^2+x+1}{{(x+3)}^2}{e}^{x-<!--\; -\; -->2}$ for $x>2$. If you'll do the sign of the first one it'll be $x^2+x-<!--\; -\; -->11<0$ therefore only $x_1=\frac{-<!--\; -\; -->1+\sqrt{45}}{2}$ verifies the conditions, for the second one it is $x^2+x-<!--\; -\; -->11>0$ therefore $x_1=\frac{-<!--\; -\; -->1+\sqrt{45}}{2}$ and $x_1=\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2}$, the last one $x^2+x+1>0$ has the solutions $x_1=\frac{-<!--\; -\; -->1-<!--\; -\; -->i\sqrt{3}}{2}$ and $x_2=\frac{-<!--\; -\; -->1+i\sqrt{3}}{2}$ (here I think that I did a mistake). Therefore our function should increase on the interval $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2})$ then decrease on the interval $(\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2},\frac{-<!--\; -\; -->1+\sqrt{45}}{2})$ then increase again on the interval $(\frac{-<!--\; -\; -->1+\sqrt{45}}{2},\frac{-<!--\; -\; -->1-<!--\; -\; -->i\sqrt{3}}{2})$ and decrease again on $(\frac{-<!--\; -\; -->1+i\sqrt{3}}{2},\mathrm{\infty <!--\; \infty \; -->})$. But Instead the solutions of the third equation are $x_1=\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{5}}{2}$ and $x_2=\frac{-<!--\; -\; -->1+\sqrt{5}}{2}$. And it is decreasing on $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2})$, then increasing on $(\frac{-<!--\; -\; -->1-<!--\; -\; -->\sqrt{45}}{2},-<!--\; -\; -->3)$ then decreasing on (-3,2) then increasing again on $(2,\mathrm{\infty <!--\; \infty \; -->})$. I don't get where I'm wrong.. neither with the results of quadratic equation nor with the intervals of monotony(sign) of the derivative.

How do you check which intervals a cubic function will increase and in which intervals it will decrease?
I was trying to find the intervals in which the cubic function $4x^3-<!--\; -\; -->6x^2-<!--\; -\; -->72x+30$ would be strictly increasing and strictly decreasing.
I managed to get the fact that at the values {-2,3} the differential of the function is zero. However this divides the function in three intervals, how can i know which intervals the function increases in and which intervals the function decreases in?
Note: I know you could simply plot the function. I was hoping for a more analytical method.

Is the floor function increasing or decreasing or both in the interval [0,1)
I was attempting a quiz in which I encountered this question, Now the graph of the floor function is constant in the interval [0,1) so technically the function should be neither increasing nor decreasing, but the answer to the question is its both increasing and decreasing. So is the answer wrong or am I unaware of a concept?
Similarly, there was another question about the function $f(x)=x^2-<!--\; -\; -->3x+2$ in the interval $[0,inf]$ but the graph of the function comes down from 0 and there is a critical point at 1 and then goes up, so is decreasing and increasing in the interval $[0,inf)$ but the answer is its neither increasing nor decreasing, now one of my classmates gave me a line of reasoning that $f^{\prime }(x)<=0and{f}^{\prime }(x)>=0$ in the inerval $[0,inf)$ which implies $f^{\prime }(x)=0$ but I'm not really convinced with the argument because the graph clearly shows that the function is decreasing then increasing. So am I missing a concept here or the answers are wrong?

Consider the function $f(x)=sup\{x^2,\cos <!--\; \; -->x\}$, $x\in <!--\; \in \; -->[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$. Show that there exists a point $x_0\in <!--\; \in \; -->[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$ such that $f(x_0)=\underset{x\in <!--\; \in \; -->[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]}{min}f(x)$. Also show that $\cos <!--\; \; -->x_0=x_0^2$.
The first part is obvious due to continuity of f in $[0,\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}]$. Second part is also obvious from the graph of f. But how to mathematically prove that $\cos <!--\; \; -->x_0=x_0^2$?
I was trying to show it using the expression of f $f(x)=\frac{1}{2}\{x^2+\cos <!--\; \; -->x+|x^2-<!--\; -\; -->\cos <!--\; \; -->x|\}$. Can anybody please help me out?

How to find where the function is decreasing/increasing/concave/convex $f(x)=\frac{2}{1+x^2}$?
I need to find where this function is increasing, decreasing, concave and convex. I've found it's derivative:
$f^{\prime }(x)=\frac{-<!--\; -\; -->4x}{(1+x^2{)}^2}$
Now you're supposed to make either $f^{\prime }(x)>0$ when it's increasing and $f^{\prime }(x)<0$ when it's decreasing, but that gives:
Increasing: $x<0$. Decreasing: $x>0$.
But what does that actually mean? It's just confusing, usually when I solve these you get 2 solutions, so it's for example increasing on the interval of (-2,2). What does this one tell me? What's the easiest way to find where this function is increasing and decreasing?
Then I also did the second derivative, which is:
$f^{″}(x)=\frac{4(3x^2-<!--\; -\; -->1)}{(1+x^2{)}^3}$
How does this all help me find my solution?

Real solution of exponential equation
1) Total no. of real solution of the equation $2^x=1+x^2$
2) Total no. of real solution of the equation $4^x=x^2$
My Try: 1) Here $x=0$ and $x=1$ are solution of Given equation.
Now we will check for other solution which are exists or not
Let $f(x)=2^x-<!--\; -\; -->x^2-<!--\; -\; -->1$, Then $f^{{}^{\prime }}(x)=2^x\ln <!--\; \; -->(2)-<!--\; -\; -->2x$ and $f^{{}^{″}}(x)=2^x{(\ln <!--\; \; -->(2))}^2-<!--\; -\; -->2$.
Now we will check the Interval of x in which function f(x) is Increasing or decreasing.
$f^{{}^{\prime }}(x)=2^x\cdot <!--\; \cdot \; -->\ln <!--\; \; -->(2)-<!--\; -\; -->2x>0$
Now I did not understand How can i proceed after that

A polynomial P(x) of degree 5 with lead coefficient one,increases in the $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)$ and $(3,\mathrm{\infty <!--\; \infty \; -->})$ and decreases in the interval (1,3)
A polynomial function P(x) of degree 5 with leading coefficient one,increases in the interval $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},1)$ and $(3,\mathrm{\infty <!--\; \infty \; -->})$ and decreases in the interval (1,3). Given that $P(0)=4$ and $P^{\prime }(2)=0$, find the value of P'(6).
I noticed that 1,2,3 are the roots of the polynomial P′(x).P′(x) is a fourth degree polynomial. So
Let $P^{\prime }(x)=(x-<!--\; -\; -->1)(x-<!--\; -\; -->2)(x-<!--\; -\; -->3)(ax+b)$.
Now I expanded P′(x) and integrated it to get P(x) and use the given condition $P(0)=4$, but I am not able to calculate a, b.

Constructing a function that is continuous and has a max on an open interval, but is not necessarily increasing immediately to the left of the max.
Suppose F(x) is continuous on some open interval I and c is a maximum point inside this interval. Is it true that f(x) must be increasing immediately to the left of c and decreasing immediately to the right of c? Proof or counterexample. (Note: A constant function is considered to be both increasing and decreasing.)

Can anyone confirm if my answer is right?
Find the critical points and the intervals of the increase and decrease of the function $f(x)=(x+5{)}^2(x-<!--\; -\; -->2{)}^5$
The critical points are: $-<!--\; -\; -->3,-<!--\; -\; -->5,2$
The four intervals -
from left to right these open intervals are $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->5),(-<!--\; -\; -->5,-<!--\; -\; -->3),(-<!--\; -\; -->3,2)\text{and}(2,\mathrm{\infty <!--\; \infty \; -->})$
This part I am not sure of:
On the interval $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},-<!--\; -\; -->5)$, f(x) is decreasing
On the interval $(-<!--\; -\; -->5,-<!--\; -\; -->3)$, f(x) is decreasing
On the interval $(-<!--\; -\; -->3,2)$, f(x) is increasing
On the interval $(2,\mathrm{\infty <!--\; \infty \; -->})$, f(x) is increasing

Derivatives and graphs
$f(x)=\sin <!--\; \; -->(x)+\cos <!--\; \; -->(x)$ for $0\le x\le 2\mathrm{\pi <!--\; \pi \; -->}$.
I have to do the following
1) Find the intervals on which f is increasing or decreasing
2) Find the local maximum and minimum values of f
3) Find the intervals of concavity and the inflections points
I got until this point when trying to solve problem 1:
$\tan <!--\; \; -->(x)=1$. The next step in the manual says that $x=\mathrm{\pi <!--\; \pi \; -->}/4$ or $5\mathrm{\pi <!--\; \pi \; -->}/4$. How did they get that $x=\mathrm{\pi <!--\; \pi \; -->}/4$ or $5\mathrm{\pi <!--\; \pi \; -->}/4$?

Rudin's PMA: Theorem 3.29 Proof
If $p>1$, $\underset{n=2}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{n(logn{)}^p}$ converges; if $p\le <!--\; \le \; -->1$, the series diverges.
Proof: The monotonicity of the logarithmic function implies that $\{logn\}$ increases. Hence $\{1/nlogn\}$ decreases, and we can apply Theorem 3.27 to the series above; this leads us to the series $\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{2}^k\cdot <!--\; \cdot \; -->\frac{1}{2^k(log{2}^k{)}^p}=\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{(klog2{)}^p}=\frac{1}{(log2{)}^p}\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\frac{1}{k^p}$ and Theorem 3.29 follows from Theorem 3.28.
I have two questions:
(1) How can we get the decrease of $\{1/nlogn\}$ from the increase of $\{logn\}$? I didn't see any conncection between them.
(2) The author said that we can apply Theorem 3.27 to the series. However, in order to apply Theorem 3.27, I think we need to show $\{1/n(logn{)}^p\}$ is decreasing. But I don't know how to do that.
Theorem 3.27: Suppose $a_1\ge <!--\; \ge \; -->a_2\ge <!--\; \ge \; -->\cdots <!--\; \cdots \; -->\ge <!--\; \ge \; -->0$. Then the series $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n$ converges if and only if the series $\underset{k=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{2}^k{a}_{2^k}=a_1+2a_2+4a_4+8a_8+\cdots <!--\; \cdots \; -->$ converges.

Is there a monotone bijection between the rationals of two intervals?
Given two nontrivial intervals I and J (both open or both closed), do there always exist a monotone bijection between $I\cap <!--\; \cap \; -->\mathbb{Q}$ and $J\cap <!--\; \cap \; -->\mathbb{Q}$?
If the endpoints of I and J are rational numbers, then such a bijection is easy to find (just take the linear function that sends the endpoints of I to those of J). But in general, it's not clear what to do.

Intervals of a derivative
I'm trying to find the intervals of the derivative of a function that I found on my book but I'm having some troubles understanding it, so I thought of trying to find some help here.
The function is $f(x)=\arcsin <!--\; \; -->(|x^2+3x+3|-<!--\; -\; -->1)$.
So I need to find the derivative of this function, study it and find the intervals (where it's increasing and where it's decreasing)
This is the solution: $-<!--\; -\; -->\frac{(2x+3)}{\sqrt{1-<!--\; -\; -->(x^2+3x+3{)}^2}}(-<!--\; -\; -->2,-<!--\; -\; -->1)$ and $\frac{(2x+3)}{\sqrt{1-<!--\; -\; -->(x^2+3x+1{)}^2}}(-<!--\; -\; -->3,-<!--\; -\; -->2)(-<!--\; -\; -->1,0)$.
So the function is decreasing in $[-<!--\; -\; -->3,-<!--\; -\; -->2]$ and $[-<!--\; -\; -->\frac{3}{2},-<!--\; -\; -->1]$ while increasing in $[-<!--\; -\; -->2,-<!--\; -\; -->\frac{3}{2}]$ and $[-<!--\; -\; -->1,0]$.
Here is where I am having some troubles:
- How do I find those intervals? I know that I need to solve the equations, but I 'm stuck there
- How can I find where the function is decreasing and increasing?

Maximmum and minimum values of function in interval
How to find the intervals ( Increasing and decreasing) of the function and its Maximum and minimum value, where the function: $f(x)=a{x}^2+bx+c$.