What steps would I take or use in order to use the intermediate value theorem to show that $\mathrm{cos}x=x$ has a solution between $x=0$ and $x=1$?

therightwomanwf
2022-07-12
Answered

What steps would I take or use in order to use the intermediate value theorem to show that $\mathrm{cos}x=x$ has a solution between $x=0$ and $x=1$?

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asked 2022-09-21

Let $h:[0,1]\to \mathbb{R}$ be continuous. Prove that there exists $w\in [0,1]$ such that

$h(w)=\frac{w+1}{2}h(0)+\frac{2w+2}{9}h\left(\frac{1}{2}\right)+\frac{w+1}{12}h(1).$

I tried several things with this problem. I first tried a new function

$\begin{array}{rl}g(x)& =h(x)-\frac{x+1}{2}h(0)+\frac{2x+2}{9}h\left(\frac{1}{2}\right)+\frac{x+1}{12}h(1)\\ & =h(x)-(x+1)(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)).\end{array}$

Evaluating $g(0)$ and $g(1)$ didn't really work.

Since I want to find a point ${x}_{0}$ where $g({x}_{0})>0$ and another ${x}_{1}$ where $g({x}_{1})<0$, I started thinking how I can make $g({x}_{0})>0$,or

$h({x}_{0})-({x}_{0}+1)(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1))>0,$

or

$\frac{h({x}_{0})}{{x}_{0}+1}>\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1).$

Since the LHS is the gradient from the point $(-1,0)$ to $({x}_{0},h({x}_{0}))$, I need to somehow prove that that

$\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)$

is always less than the steepest gradient from $(-1,0)$ to $({x}_{0},h({x}_{0}))$. However, I am stuck here. Any help?

$h(w)=\frac{w+1}{2}h(0)+\frac{2w+2}{9}h\left(\frac{1}{2}\right)+\frac{w+1}{12}h(1).$

I tried several things with this problem. I first tried a new function

$\begin{array}{rl}g(x)& =h(x)-\frac{x+1}{2}h(0)+\frac{2x+2}{9}h\left(\frac{1}{2}\right)+\frac{x+1}{12}h(1)\\ & =h(x)-(x+1)(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)).\end{array}$

Evaluating $g(0)$ and $g(1)$ didn't really work.

Since I want to find a point ${x}_{0}$ where $g({x}_{0})>0$ and another ${x}_{1}$ where $g({x}_{1})<0$, I started thinking how I can make $g({x}_{0})>0$,or

$h({x}_{0})-({x}_{0}+1)(\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1))>0,$

or

$\frac{h({x}_{0})}{{x}_{0}+1}>\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1).$

Since the LHS is the gradient from the point $(-1,0)$ to $({x}_{0},h({x}_{0}))$, I need to somehow prove that that

$\frac{1}{2}h(0)+\frac{2}{9}h\left(\frac{1}{2}\right)+\frac{1}{12}h(1)$

is always less than the steepest gradient from $(-1,0)$ to $({x}_{0},h({x}_{0}))$. However, I am stuck here. Any help?

asked 2022-09-25

Let $f\in {L}^{1}({\mathbb{R}}^{2})$ with respect to the Lebesgue measure $m\times m$ on ${\mathbb{R}}^{2}$. Prove that if

${\iint}_{{\mathbb{R}}^{2}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0,$

then there exits a square ${S}_{a,b}=\{(x,y)\mid a\le x\le a+1,b\le y\le b+1\}$, such that

${\iint}_{{S}_{a,b}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0.$

I tried to show that the integral

${\iint}_{[a,a+x]\times [b,b+y]}f(s,t)\phantom{\rule{thinmathspace}{0ex}}ds\phantom{\rule{thinmathspace}{0ex}}dt$

is absolutely continuous by Fubini's Theorem and Fundamental Theorem. And by the countable additivity of integration, I proved the integral on the whole plane is still A.C. However, I could not directly apply a theorem like the IVT for the single variable functions.

Is there any theorem for the two-dimensional case?

${\iint}_{{\mathbb{R}}^{2}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0,$

then there exits a square ${S}_{a,b}=\{(x,y)\mid a\le x\le a+1,b\le y\le b+1\}$, such that

${\iint}_{{S}_{a,b}}f(x,y)\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy=0.$

I tried to show that the integral

${\iint}_{[a,a+x]\times [b,b+y]}f(s,t)\phantom{\rule{thinmathspace}{0ex}}ds\phantom{\rule{thinmathspace}{0ex}}dt$

is absolutely continuous by Fubini's Theorem and Fundamental Theorem. And by the countable additivity of integration, I proved the integral on the whole plane is still A.C. However, I could not directly apply a theorem like the IVT for the single variable functions.

Is there any theorem for the two-dimensional case?

asked 2022-04-07

Use the Intermediate Value Theorem to prove that the equation

$-{x}^{3}+4\mathrm{sin}(x)+4{\mathrm{cos}}^{2}(x)=0$

has at least two solutions. You should carefully justify each of the hypothesis of the theorem.

How do I know that at least two exist because I know you're meant to look for a change in sign.

$-{x}^{3}+4\mathrm{sin}(x)+4{\mathrm{cos}}^{2}(x)=0$

has at least two solutions. You should carefully justify each of the hypothesis of the theorem.

How do I know that at least two exist because I know you're meant to look for a change in sign.

asked 2022-05-21

I'm aware that $\mathbb{R}$ is not the only set that satisfies the least upper bound property, the $p$-adics do also. Does the intermediate value theorem also hold in the $p$-adics then?

What are the spaces where the intermediate value theorem hold?

What are the spaces where the intermediate value theorem hold?

asked 2022-08-11

Let's consider the continuous function

$f:\mathbb{R}\times [a,b]\to \mathbb{R}$

Such that $f(x,a)>0$ for all $x$ and there exists ${x}_{b}$ with $f({x}_{b},b)\le 0$.

Then there exists $c\in [a,b]$ such that $f(x,c)\ge 0$ 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.

$f:\mathbb{R}\times [a,b]\to \mathbb{R}$

Such that $f(x,a)>0$ for all $x$ and there exists ${x}_{b}$ with $f({x}_{b},b)\le 0$.

Then there exists $c\in [a,b]$ such that $f(x,c)\ge 0$ 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.

asked 2022-09-22

Use the intermediate value theorem to prove that if $f:[0,1]\to [0,1]$ is continuous, then there exists $c\in [0,1]$ such that $f(c)=\sqrt{c}$

Suppose that $f(0)<f(1)$. Consider now, $f(0)<\sqrt{c}<f(1)$. By the intermediate value theorem, $\mathrm{\exists}b\in [0,1]:f(b)=\sqrt{c}$.

Now we need to show that its possible that $b=c$, how?

Suppose that $f(0)<f(1)$. Consider now, $f(0)<\sqrt{c}<f(1)$. By the intermediate value theorem, $\mathrm{\exists}b\in [0,1]:f(b)=\sqrt{c}$.

Now we need to show that its possible that $b=c$, how?

asked 2022-07-19

Use the intermediate value theorem to show that the equation,

$tan(x)=2x$

has an infinite amount of real solutions.

So far I have used the IVT to show that for $f(x)=tan(x)$ in the interval $(-\frac{\pi}{2},\frac{\pi}{2})$ there is a $L$ between $-\mathrm{\infty}$ and $\mathrm{\infty}$ such that there is a value $c$ where $f(c)=L$

I also have shown that for $g(x)=2x$ in the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ there is a $K$ between $-\pi $ and $\pi $ such that there is a value $d$ where $f(d)=K$

Would it be correct to say that because each of these functions have a value for each point in the interval and that the range of $f(x)$ consists of all reals that $f(x)$ must intersect with $g(x)$ or there is some value $x$ in the interval such that $2x=tan(x)$? If this is correct then how would I extend this result to all intervals $[\pi (n-\frac{1}{2}),\pi (n+\frac{1}{2})]$?

$tan(x)=2x$

has an infinite amount of real solutions.

So far I have used the IVT to show that for $f(x)=tan(x)$ in the interval $(-\frac{\pi}{2},\frac{\pi}{2})$ there is a $L$ between $-\mathrm{\infty}$ and $\mathrm{\infty}$ such that there is a value $c$ where $f(c)=L$

I also have shown that for $g(x)=2x$ in the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$ there is a $K$ between $-\pi $ and $\pi $ such that there is a value $d$ where $f(d)=K$

Would it be correct to say that because each of these functions have a value for each point in the interval and that the range of $f(x)$ consists of all reals that $f(x)$ must intersect with $g(x)$ or there is some value $x$ in the interval such that $2x=tan(x)$? If this is correct then how would I extend this result to all intervals $[\pi (n-\frac{1}{2}),\pi (n+\frac{1}{2})]$?