Prove that there exists a real number $x$ such that ${x}^{177}+\frac{165}{1+{x}^{8}+{\mathrm{sin}}^{2}(x)}=125$ using Intermediate Value Theorem.

Josiah Owens
2022-10-26
Answered

Prove that there exists a real number $x$ such that ${x}^{177}+\frac{165}{1+{x}^{8}+{\mathrm{sin}}^{2}(x)}=125$ using Intermediate Value Theorem.

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Remington Wells

Answered 2022-10-27
Author has **13** answers

Define

$f(x)={x}^{177}+\frac{165}{1+{x}^{8}+{\mathrm{sin}}^{2}(x)}-125.$

Clearly $f(x)$ is continuous in $[-2,2]$. Since $f(-2)<0,f(2)>0$, there is $c\in (-2,2)$ such that $f(c)=0$.

$f(x)={x}^{177}+\frac{165}{1+{x}^{8}+{\mathrm{sin}}^{2}(x)}-125.$

Clearly $f(x)$ is continuous in $[-2,2]$. Since $f(-2)<0,f(2)>0$, there is $c\in (-2,2)$ such that $f(c)=0$.

asked 2022-07-21

$a,b>0$ and $c<d$

Show that $g(x)={\displaystyle \frac{a}{(x-c{)}^{4}}+\frac{b}{(x-d{)}^{7}}=0}$ has at least a solution in $(c,d)$

I defined

$\underset{x\searrow c}{lim}g(x)=+\mathrm{\infty}$

$\underset{x\nearrow d}{lim}g(x)=-\mathrm{\infty}$

I don't know if the limits are just enough for the proof, or is there something else?

Show that $g(x)={\displaystyle \frac{a}{(x-c{)}^{4}}+\frac{b}{(x-d{)}^{7}}=0}$ has at least a solution in $(c,d)$

I defined

$\underset{x\searrow c}{lim}g(x)=+\mathrm{\infty}$

$\underset{x\nearrow d}{lim}g(x)=-\mathrm{\infty}$

I don't know if the limits are just enough for the proof, or is there something else?

asked 2022-09-21

I know I have to use the intermediate value theorem but how?

asked 2022-09-20

Question: Sketch all the continuous functions $f:\mathbb{R}\to \mathbb{R}$ which satisfy

$(f(x){)}^{2}=(x-1{)}^{2}(x-2{)}^{2}.$

Justify your answers.

I have found the eight possible continuous functions as follows:

$\begin{array}{rl}f(x)& =(x-1)(x-2)\\ f(x)& =-(x-1)(x-2)\\ f(x)& =|x-1|(x-2)\\ f(x)& =-|x-1|(x-2)\\ f(x)& =(x-1)|x-2|\\ f(x)& =-(x-1)|x-2|\\ f(x)& =|x-1||x-2|\\ f(x)& =-|x-1||x-2|.\end{array}$

and have got the conclusion of this question which has at most eight possible such functions, but how to proof the "at most" statement.

$(f(x){)}^{2}=(x-1{)}^{2}(x-2{)}^{2}.$

Justify your answers.

I have found the eight possible continuous functions as follows:

$\begin{array}{rl}f(x)& =(x-1)(x-2)\\ f(x)& =-(x-1)(x-2)\\ f(x)& =|x-1|(x-2)\\ f(x)& =-|x-1|(x-2)\\ f(x)& =(x-1)|x-2|\\ f(x)& =-(x-1)|x-2|\\ f(x)& =|x-1||x-2|\\ f(x)& =-|x-1||x-2|.\end{array}$

and have got the conclusion of this question which has at most eight possible such functions, but how to proof the "at most" statement.

asked 2022-08-15

$f(x)={x}^{2}\mathrm{exp}(\mathrm{sin}(x))-\mathrm{cos}(x)$ on the interval $[0,\pi /2]$, I have shown the function is continuous and that there is at least one solution on the interval via using IVT, I know I have to find another solution in the interval such that $f(x)<0$, $f(x)>0$ and then $f(x)<0$, just wondering what is the best way to approach, Im sure there must be a more effective way than just to number crunch, could we consider MVT on the interval, many thanks in advance.

asked 2022-08-22

$f$ is continuous function in $[0,2]$ and $f(2)=1$

proof that there is a point $x$ in $[0,2]$ so that:

$f(x)=\frac{1}{x}$

nobody in my class solved this exercise. our teacher give us a hint: $p(x)=x\ast f(x)$ and to do Intermediate value theorem on this function. so I tried:

$p(0)=0\ast f(0)=0,P(2)=2\ast f(2)=2$

$p(0)<h<p(2)$ by Intermediate value theorem, there is a point $c$ so that $a<c<b$ and $f(c)=h$

please solve

proof that there is a point $x$ in $[0,2]$ so that:

$f(x)=\frac{1}{x}$

nobody in my class solved this exercise. our teacher give us a hint: $p(x)=x\ast f(x)$ and to do Intermediate value theorem on this function. so I tried:

$p(0)=0\ast f(0)=0,P(2)=2\ast f(2)=2$

$p(0)<h<p(2)$ by Intermediate value theorem, there is a point $c$ so that $a<c<b$ and $f(c)=h$

please solve

asked 2022-07-12

How would I go about showing this question:

Suppose that the function $f:[0,1]\to [0,1]$ is continuous. Use the Intermediate Value Theorem to prove that there exists $c\in [0,1]$ such that

$f(c)=c(2-{c}^{2})$

My thoughts are we can define a function say $g(x):=f(x)-x(2-{x}^{2})$ and then $g(0)=f(0)$ and then $g(1)=f(1)-1$ and then I'm stuck on how to complete it to get to the point where

$g(0)<0<g(1)$

so there would exist a c s.t. $g(c)=0$ [by IVT] and thus $f(c)=c(2-{c}^{2})$

If anyone could help, would be much appreciated thank you!

Suppose that the function $f:[0,1]\to [0,1]$ is continuous. Use the Intermediate Value Theorem to prove that there exists $c\in [0,1]$ such that

$f(c)=c(2-{c}^{2})$

My thoughts are we can define a function say $g(x):=f(x)-x(2-{x}^{2})$ and then $g(0)=f(0)$ and then $g(1)=f(1)-1$ and then I'm stuck on how to complete it to get to the point where

$g(0)<0<g(1)$

so there would exist a c s.t. $g(c)=0$ [by IVT] and thus $f(c)=c(2-{c}^{2})$

If anyone could help, would be much appreciated thank you!

asked 2022-09-29

Show that $\sqrt{1+x}<1+\frac{x}{2}$ if $x>0$