Help to Integral Calculus Solutions & Examples

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.

1. Write this second order ode as a first order ode.
$y^{″}(t)=y(t)+y^{\prime }(t)+t$
$y^{\prime }(0)=1$
$y^{\prime }(0)=0$
Use the Euler Method with h = 0.1. What approximation of $(y(h),y^{\prime }(h))$ do you get?"
I think I understand the first problem. I substitute y with $u_1$ and $y^{\prime }$ with $u_2$ which gives: $y^{\prime }=u_2,y^{″}=u_1+u_2+t$. Is that correct?
If so, I don't really know where to go from there.

Trying to understand the proof that the numerical method of solving differential equation
$x_{i+1}=x_i+{\mathrm{\tau <!--\; \tau \; -->}}_{i}F(t_i+\frac{{\mathrm{\tau <!--\; \tau \; -->}}_i}{2},x_{i+\frac{1}{2}})$
$x_{i+\frac{1}{2}}=x_i+\frac{\mathrm{\tau <!--\; \tau \; -->}}{2}F(t_i,x_i)$
which seems to be the Cauchy-Euler's method.
I am stuck with the Taylor approximation of
$x(t+\mathrm{\tau <!--\; \tau \; -->})=x(t)+\mathrm{\tau <!--\; \tau \; -->}{x}^{\prime }(t)+\frac{1}{2}{x}^{″}(t)+O({\mathrm{\tau <!--\; \tau \; -->}}^3)$
Here it is OK for me.
I also understand, why $x^{\prime }(t)=F(t,x(t))$.
But why
$x^{″}(t)=\frac{d}{dt}F(t,x(t))={\mathrm{\partial <!--\; \partial \; -->}}_{t}F(t,x(t))+D_{x}F(t,x(t))F(t,x(t))$???
I am a little frustrated about this. Normally I am more into computer science and some mathematical concepts can be missing, so sorry for possible trivial or illposed question, thanks for patience :)

I have shown that for the given ODE system, that when we apply the forward Euler method to something like
$\begin{array}{rl}{\mathbf{y}}^{\prime }& =A\mathbf{y}\\ \mathbf{y}(t_0)& =y_0\\ t& \in <!--\; \in \; -->(t_0,T]\end{array}$
where $A$ is diagonalizable; $A=PD{P}^{-<!--\; -\; -->1}$, that the solution given by forward Euler is
$\begin{array}{r}{u}_n=P(I+hD{)}^n{P}^{-<!--\; -\; -->1}{u}_0\end{array}$
where we consider the Forward Euler method to be $y_{k+1}=y_k+h{y}^{\prime }(x_k,y_k).$ N
Now, I want to show that if the real parts of ${\mathrm{\lambda <!--\; \lambda \; -->}}_k$ (the eigenvalues of $A$ and thus the entries of $D$) are negative, then $h<\frac{-<!--\; -\; -->2Re{\mathrm{\lambda <!--\; \lambda \; -->}}_k}{|{\mathrm{\lambda <!--\; \lambda \; -->}}_k{|}^2}$ for all $k$ implies that $u_n\to <!--\; \to \; -->0$.
Now, in order to show this, based on what we have obtained for the formula for $u_n$, we need only show that for each $|1+h{\mathrm{\lambda <!--\; \lambda \; -->}}_k|<1$. I am having a really tough time doing this, and have mostly just tried to play around with AM-GM stuff.

For standard normal random variable Z, determine the value of constant c which makes the probability statements given below correct
a) $\mathrm{\Phi <!--\; \Phi \; -->}(c)=0.8888$
b) $P(0\le <!--\; \le \; -->Z\le <!--\; \le \; -->c)=0.334$
c) $P(-<!--\; -\; -->c\le <!--\; \le \; -->Z\le <!--\; \le \; -->c)=0.488$
e) $P(c\le <!--\; \le \; -->|Z|)=0.154$

Use Euler's method to find approximate values for the solution of the initial value − problem
$\frac{dy}{dx}=x-<!--\; -\; -->y$
$y(0)=1$
on the interval [0,1] using five steps of size h=0.2.

My attempts:
I know that the recurrence relation $y_{n+1}=y_n+hf(x_n,y_n)$ however I am unable to see how the interval comes into play.An idea I had was to consider the bounds of the interval and approximate y(0) and y(1) however this does not include h so I am extremely skeptical.

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.

Finding total number of asymptotes
I was solving some problems Related to asymptotes, I have a question Can we know number of asymptotes of a 1.rational function 2.trignometric function 3. Implicit function
Without calculating them how can we know how many total number of asymptotes function will have? is we guess this from degree of denominator? but in implicit functions we cannot separate y to put in fraction? plz help me

Graphic intersecting asymptotes
Sometimes graphics intersect the asymptotes(horizontal) of the function we plot and then they tend to the asymptote to infinity.What gives us the information whether the graph only tends to the asymptote and does not intersect it or intersects the asymptote a then tends to her for a given point?

I am having trouble to convert 1.8768686868... or 0.95287928792879...
If there is a quick simple technique for this?

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 $\mathrm{\alpha <!--\; \alpha \; -->},$

Asymptotes of $\arctan <!--\; \; -->2x$
My book tells me the horizontal asymptotes of $\arctan <!--\; \; -->2x$ is either at positive or negative $\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$, yet the vertical asymptotes of $\tan <!--\; \; -->2x$ occurs at positive or negative $x=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$, so obviously the horizontal asymptotes of $\arctan <!--\; \; -->2x$ should be at either positive or negative $y=\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$.
Can someone tell me what I'm doing wrong?

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?

I'm writing a MATLAB function to solve a system of differential equations with the Euler method. I need to estimate the local truncation error of the method but I don't know how to do.
I know that the local truncation error of a first order DE $y^{\prime }=f(y(t),t)$ is given by :
$LT{E}_i=-<!--\; -\; -->\frac{h^2}{2}{y}^{″}({\mathrm{\xi <!--\; \xi \; -->}}_i)$
where $h$ is the step size and ${\mathrm{\xi <!--\; \xi \; -->}}_i\in <!--\; \in \; -->[t_{i-<!--\; -\; -->1},t_i]$.
What is the expression for the local truncation error of the system $\stackrel{\to <!--\; \to \; -->}{y^{\prime }}=f(\stackrel{\to <!--\; \to \; -->}{y}(t),t)$?

What is the remainder when ${17}^{100000000\dots <!--\; \dots \; -->}$, with $n$ zeroes is divided by 20? I.e.:
${17}^{{10}^n}\mathrm{mod}20=?$
My approach: I can solve it by Euler's totient function ; ${17}^8=1\mathrm{mod}20$, so remainder is 1.
But I want to know if the following method will fetch the result: 17 divided by 20 gives −3 as remainder and −3 raised to any multiple of $10\mathrm{mod}20$ gives 1?
Is this logic correct? I think it works only for positive remainder.

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?

I'm solving a system of stiff ODEs, at first I wanted to implement BDF, but it seem to be a quite complicated method, so I decided to start with Backward Euler method. Basically it says that you can solve an ODE:
$y^{\prime }=f(t,y)$, with initial guess $y(t_0)=y_0$
Using the following approximation:
$y_{k+1}=y_k+hf(t_{k+1},y_{k+1})$, where $h$ is a step size on parameter $t$
Wikipedia article says that you can solve this equation using Newton-Raphson method, which is basically a following iteration:
$x_{n+1}=x_n-<!--\; -\; -->\frac{g(x_n)}{g^{\prime }(x_n)}$
So, the question is how to correctly mix them together? What initial guess $x_0$ and function $g$ should be?
Also $f$ is quite complex in my case and I'm not sure if it possible to find another derivative of it analytically. I want to write an implementation of it by myself, so pblackefined Mathematica functions wouldn't work.