Master Intermediate Value Theorem Problems

Let $f(x)=x^5-<!--\; -\; -->5x+p$. Show that $f$ can have at most one root in $[0,1]$,regardless of the value of p.

This seems to be an IVT problem, so I will go forth with that:

$f(a)><!--\; >\; -->u><!--\; >\; -->f(b)$
$f(0)><!--\; >\; -->0><!--\; >\; -->f(1)$
$p><!--\; >\; -->0><!--\; >\; -->p-<!--\; -\; -->4$
This is only true for $p:(0,4)$ ie $0<<!--\; <\; -->p<<!--\; <\; -->4$
Now I have proved that there is atleast one root between $[0,1]$, now I need to prove that there isn't a second.

Would Rolles theorem be what I need here, or is there a simple deduction I can make from above to finish the problem?

You've got a job at a company and you will be traveling to a conference with accommodation. You leave home on monday at 07:15 and arrive at 11:07. The next day you drive exactly the same route back. There is less traffic so you start at 8:32 and arrive at 11:01. Show that there is a point on that route you were at, at the exact same time the two days.

Okay, so that is my question. Does anyone know how to go about this one? I think I need to use the intermediate value theorem, but I'm not sure how to! Thanks in advance for tips/solutions.

a) $e^x=2-<!--\; -\; -->x^2$
b) $\sin <!--\; \; -->(x)=x^2-<!--\; -\; -->1$
The intermediate value theorem says if $F$ is a continues function in a closed interval $[A,B]$, and you choose a value $K$ between $f(A)$ and $f(B)$ where $f(A)\ne f(B)$, Then, there is a value between $A$ and $B$, $C$, such that $f(C)=K$.

I tried to solve the first one by trying to do what the intermediate value theorem says. Then I went on by choosing 2 values A and B. But the first equation has a weird behavior and so does the second one.

Their graphs have 2 long vertical lines with a medium space between them, and Choosing values to plug in doesn't return exact numbers, I'm confused, if anyone can give more clarifications about these problems it would be of great help for me.

Given $f:[0,1]\to <!--\; \to \; -->{\mathbb{R}}^2$ such that $f(0)=(1,0),f(1)=(-<!--\; -\; -->1,0)$ and $f$ is continuous with respect to the standard topologies. How can I prove that there exists $x\in <!--\; \in \; -->[0,1]$ such that $f(x)=(0,y)$, for some $y\in <!--\; \in \; -->\mathbb{R}$?
My attempt to a solution was to consider the restriction of $f$ to the set
$A=\{x:f(x)=(y,0)\text{for some}y\in <!--\; \in \; -->\mathbb{R}\}\subset <!--\; \subset \; -->[0,1].$
Since $f$ is continuous then its restriction is continuous.
Since the restriction of $f$ to $A$ is "basically" a map from $A$ to $\mathbb{R}$, it makes sense that I would just need to apply IVT to finish the proof. I'm having issue in formalizing the reasoning that the restriction is "basically" a map from $A$ to $\mathbb{R}$.

Let's consider the continuous function
$f:\mathbb{R}\times <!--\; \times \; -->[a,b]\to <!--\; \to \; -->\mathbb{R}$
Such that $f(x,a)>0$ for all $x$ and there exists $x_b$ with $f(x_b,b)\le <!--\; \le \; -->0$.
Then there exists $c\in <!--\; \in \; -->[a,b]$ such that $f(x,c)\ge <!--\; \ge \; -->0$ for all $x$ and a real value $x_c$c with $f(x_c,c)=0$.
Intuitively this seems true, but I woludn't know how to prove it.

IVP: A function $f$ has the intermediate value property on an interval $[a,b]$ if for all $x<y$ in $[a,b]$ and all $L$ between $f(x)$ and $f(y)$, it is always possible to find a point $c\in <!--\; \in \; -->(x,y)$ where $f(c)=L$

And

IVT: If $f:[a,b]\to <!--\; \to \; -->\mathbb{R}$ is continuous, and if $L$ is a real number satisfying $f(a)<L<f(b)$ or $f(a)>L>f(b)$ then there exists a point $c\in <!--\; \in \; -->(a,b)$ where $f(c)=L$Both definitions seem very similar to me, what purpose do they serve individually?

Let $f$ and $g$ be continuous functions on $[a,b]$ such that $f(a)\ge <!--\; \ge \; -->g(a)$ and $f(b)\le <!--\; \le \; -->g(b)$. Prove $f(x_0)=g(x_0)$ for at least one $x_0$ in $[a,b]$.
Here's what I have so far:
Let $h$ be a continuous function where $h=f-<!--\; -\; -->g$
Since $h(b)=f(b)-<!--\; -\; -->g(b)\le <!--\; \le \; -->0$, and $h(a)=f(a)-<!--\; -\; -->g(a)\ge <!--\; \ge \; -->0$, then $h(b)\le <!--\; \le \; -->0\le <!--\; \le \; -->h(a)$ is true.
So, by IVT there exists some $y$ such that...
And that's where I need help. I read what the IVT is, but I could use some help explaining how it applies here, and why it finishes the proof. Thank you!

Intermediate value theorem: show there exists $c\in <!--\; \in \; -->[0,2/3]$ such that $f(c+\frac{1}{3})=f(c)$
Let $f:[0,1]\to <!--\; \to \; -->[0,1]$ be continuous and $f(0)=f(1)$
This is pretty straight forward and the other ones ive done it was easier to find why IVT applies, namely find why it changes signs.
Letting $g(x)=f(x+\frac{1}{3})-<!--\; -\; -->f(x)$ then this is only defined for the interval $[0,\frac{2}{3}]$ b/c not sure what $g(1)=f(\frac{4}{3})-<!--\; -\; -->f(1)$ may be.
Now, $g(0)=f(\frac{1}{3})-<!--\; -\; -->f(0)$
and $g(\frac{2}{3})=f(1)-<!--\; -\; -->f(\frac{2}{3})=f(0)-<!--\; -\; -->f(\frac{2}{3})$.
I can't (at least I don't see why I can) conclude this is equal to $-<!--\; -\; -->g(0)$ which has been the case in other exercises, which is what I used to apply the IVT and get the wanted result.

Suppose $f:[-<!--\; -\; -->1,1]\to <!--\; \to \; -->\mathbb{R}$ is continuous, $f(-<!--\; -\; -->1)>-<!--\; -\; -->1$, and $f(1)<1$. Show that $f$ has a fixed point.
So for every $y$ between $f(-<!--\; -\; -->1)$ and $f(1)$ we can find an $x$ s.t. $-<!--\; -\; -->1<x<1$ and $f(x)=y$, but I'm not sure how to show one of these creates a fixed point.

Theorem: Let $f$ be continuous on $[a,b]$ and assume $f(a)\ne f(b)$. Then for every $\mathrm{\lambda <!--\; \lambda \; -->}$ such that $f(a)<\mathrm{\lambda <!--\; \lambda \; -->}<f(b)$, there exists a $c\in <!--\; \in \; -->(a,b)$ such that $f(c)=\mathrm{\lambda <!--\; \lambda \; -->}$.

Question:
Suppose that $f:[0,1]\to <!--\; \to \; -->[0,2]$ is continuous. Use the Intermediate Value Theorem to prove that their exists $c\in <!--\; \in \; -->[0,1]$ such that:
$f(c)=2c^2$

Attempt:
I know that when we have the condition were $f:[a,b]\to <!--\; \to \; -->[a,b]$, the method to prove that c exits, is the same method you would use to prove the fixed point theorem.

Unfortunately I don't have an example in my notes when we have $f:[a,b]\to <!--\; \to \; -->[a,y]$. How would I use the IVT to answer the original question?

Let $f(x)={\displaystyle \frac{20}{(x^6+x^4+x^2+1)}}$.
I need to show that for any $k\in <!--\; \in \; -->[5,20]$, there is a point $c\in <!--\; \in \; -->[0,1]$ such that $f(c)=k$.
Thanks in advance.

Prove that the polynomial $x^6+x^4-<!--\; -\; -->5x^2+1$ has at least four real roots.
Talking analysis here, using the definition of continuity, intermediate value theorem, and extreme value theorem.

I have so far written the following proof in terms I understand for the Mean Value Theorem for Integrals:

'If $f$ is continuous on an interval $I=[a,b]$, then for some $c\in <!--\; \in \; -->I$ we have:
${\int <!--\; \int \; -->}_a^{b}f(x)\cdot <!--\; \cdot \; -->dx=f(c)(b-<!--\; -\; -->a)$
Let $m$ and $M$ denote the minimum and maximum values of $f$ on $I$. Then: $m\le <!--\; \le \; -->f(x)\le <!--\; \le \; -->M$ for all $x\in <!--\; \in \; -->[a,b]$. Integrating the inequalities we should find that:
${\int <!--\; \int \; -->}_a^{b}m\le <!--\; \le \; -->{\int <!--\; \int \; -->}_a^{b}f(x)\le <!--\; \le \; -->{\int <!--\; \int \; -->}_a^{b}M$
$m(b-<!--\; -\; -->a)\le <!--\; \le \; -->{\int <!--\; \int \; -->}_a^{b}f(x)\le <!--\; \le \; -->M(b-<!--\; -\; -->a)$
$m\le <!--\; \le \; -->\frac{1}{b-<!--\; -\; -->a}{\int <!--\; \int \; -->}_a^{b}f(x)\le <!--\; \le \; -->M$
However, the book I am using now says: 'label
$\frac{1}{b-<!--\; -\; -->a}{\int <!--\; \int \; -->}_a^{b}f(x)$
as $A(f)$. But now the intermediate value theorem says $A(f)=f(c)$ for some $c\in <!--\; \in \; -->I$. I am confused how the intermediate value theorem shows this and was wondering if anyone could help me understand or clarify it?

Evaluate each expression for the given values
-k to second power -(3k-5n) + 4n when k= -1 and n = -2
please show your work

(d) Let $f:[a,b]\to <!--\; \to \; -->[a,b]$ be a continuous function. Prove that $f$ has a fixed point, i.e. that there is a $с\in <!--\; \in \; -->[a,b]$ such that $f(c)=c$.
In the solution, it says that $f(a)\ge <!--\; \ge \; -->a$ and $f(b)\le <!--\; \le \; -->b$ but it do not seem obvious for me. If I am just given that a $f:[a,b]\to <!--\; \to \; -->[a,b]$, how do I know is this function increasing, decreasing or just a horizontal line?

I am not sure if the IVT should be applied here. I am try to do this problem and am stuck an how proceed:
Suppose $f:[-<!--\; -\; -->1,1]\to <!--\; \to \; -->\mathbb{R}$ is continuous and satisfies $f(-<!--\; -\; -->1)=f(1)$. Prove that there exists $y\in <!--\; \in \; -->[0,1]$ such that $f(y)=f(y-<!--\; -\; -->1)$.
So far, I have consideblack a new function $g(x)=f(x)-<!--\; -\; -->f(x-<!--\; -\; -->1)$, $x\in <!--\; \in \; -->[0,1]$, but am stuck after this.

The temperature $T(x)$ at each point $x$ on the surface of Mars (a sphere) is a continuous function. Show that there is a point $x$ on the surface such that $T(x)=T(-<!--\; -\; -->x)$
(Hint: Represent the surface of Mars as $\{x\in <!--\; \in \; -->{\mathbb{R}}^3:||x||=1\}$.)
Consider the function $f(x)=T(x)-<!--\; -\; -->T(-<!--\; -\; -->x)$
So.....
I consider an unit sphere is locating at the origin of a $xyz$-plane.
As $||x||=1$, I can say with $radius=1=\sqrt{x^2+y^2+z^2}$
To find there is a point $T(x)=T(-<!--\; -\; -->x)$, we use the formula $f(x)=T(x)-<!--\; -\; -->T(-<!--\; -\; -->x)$ and show somehow $f(x)$ will equal to 0 ???
It will be a point in the upper hemisphere and another point with the exactly opposite vector (if using $ijk$ plane) on the lower hemisphere

Suppose that $x$ is a solution to the initial value problem $x^{\prime }=\frac{x^3-<!--\; -\; -->x}{1+t^2{x}^2}$ and $x(0)=\frac{1}{2}$ Show that $0<x(t)<1$ for all $t$ for which $x$ is defined.
For this question, I am told to use proof by contradiction. First find constant solutions, and then use intermediate value theorem to argue that since constant solution and initial value solution don't cross, by uniqueness theorem their existence contradict each other, hence $0<x(t)<1$. Can anyone explain how intermediate value theorem plays an effect in here?

$f(x$) = $x^5-<!--\; -\; -->15x^4-<!--\; -\; -->10x^2+20$
Prove that the equation $f(x)=0$ has a real solution with $x\ge <!--\; \ge \; -->1$.
Idea: Started off by coming up with a boundary $[1,x]$ for some real number $x$. $f(1)=-<!--\; -\; -->4$ so is less than zero, then attempted to find an x that gave me a positive result, very quickly came to the conclusion that any number above 1 when inserted into a function gave me a negative number. Pretty sure I could be missing something really simple. Any help or hints would be appreciated.