Continuity and limts: Show a function is continuous at a

$f\left(x\right)={(x+2{x}^{3})}^{4},a=-1$

nagasenaz
2021-10-29
Answered

Continuity and limts: Show a function is continuous at a

$f\left(x\right)={(x+2{x}^{3})}^{4},a=-1$

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Sadie Eaton

Answered 2021-10-30
Author has **104** answers

Show a function is continuous at a

$f\left(x\right)={(x+2{x}^{3})}^{4},a=-1$

We have$f\left(x\right)={(x+2{x}^{3})}^{4},a=-1$

$f\left(x\right)={(x+2{x}^{3})}^{4}$ at a=-1 to be continous

if, L.H.L=R.H.L=Function at the given point

Now,

$\underset{x\to -1}{lim}f\left(x\right)=\underset{x\to -1}{lim}{(x+2{x}^{3})}^{4}={(\underset{x\to -1}{lim}x+2{\left(\underset{x\to -1}{lim}x\right)}^{3})}^{4}$

$\Rightarrow \underset{x\to -1}{lim}f\left(x\right)={(-1+2{(-1)}^{3})}^{4}={(-1-2)}^{4}={(-3)}^{4}$

$\Rightarrow \underset{x\to -1}{lim}f\left(x\right)=81$

$\therefore L.H.L=\underset{x\to -1}{lim}f\left(x\right)=81$

Then,

$\underset{x\to 1}{lim}f\left(x\right)={(1+2{\left(1\right)}^{3})}^{5}={(1+2)}^{4}={\left(3\right)}^{4}$

$\therefore R.H.L=\underset{x\to 1}{lim}f\left(x\right)=81$

Now, for f(-1)

$\Rightarrow f(-1)={((-1)+2{(-1)}^{3})}^{4}={(-1-2)}^{4}={(-3)}^{4}$

$\Rightarrow f(-1)=81$

Here,$L.H.L=R.H.L=f(-1)$

Hence, f(x) is continuous at a=-1

We have

if, L.H.L=R.H.L=Function at the given point

Now,

Then,

Now, for f(-1)

Here,

Hence, f(x) is continuous at a=-1

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Notation: If $f:\mathbb{R}\to \mathbb{R}$ is continuous, let us denote $If:\mathbb{R}\to \mathbb{R}$ its indefinite integral from 0, i.e., $(If)(x)={\int}_{0}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$, and iteratively ${I}^{k+1}f=I({I}^{k}f)$.

Remark: If f is a continuous function with support contained in the open interval ]0,1[ then If has support contained in ]0,1[ iff $(If)(1)=0$.

Main question: Does there exist a ${C}^{\mathrm{\infty}}$ function f with support contained in the open interval ]0,1[ such that ${I}^{k}f$ has support contained in ]0,1[ for every $k\ge 0$, or, equivalently, $({I}^{k}f)(1)=0$ for all $k\ge 0$?

Equivalent formulation: Does there exists a sequence $({f}_{k}{)}_{k\in \mathbb{Z}}$ of ${C}^{\mathrm{\infty}}$ functions each with support contained in the open interval ]0,1[, such that ${f}_{k-1}$ is the derivative of ${f}_{k}$?

Weaker question: Does there at least exist a continuous function f with the properties demanded in the main question?

Stronger question: Does there exist a ${C}^{\mathrm{\infty}}$ function f with compact support, whose Fourier transform vanishes identically on a nontrivial interval?

(A positive answer to the latter would imply a positive answer to the main question: rescale the function so its support is contained in ]0,1[, multiply it appropriately so its Fourier transform vanishes in a neighborhood of 0, and observe that the Fourier transform of ${I}^{k}f$ is, up to constants, ${\xi}^{k}$ times that of f.)

Edit: Before someone points out that the identically zero function fits the bill, I should add that I want my functions to not vanish identically.

Notation: If $f:\mathbb{R}\to \mathbb{R}$ is continuous, let us denote $If:\mathbb{R}\to \mathbb{R}$ its indefinite integral from 0, i.e., $(If)(x)={\int}_{0}^{x}f(t)\phantom{\rule{thinmathspace}{0ex}}dt$, and iteratively ${I}^{k+1}f=I({I}^{k}f)$.

Remark: If f is a continuous function with support contained in the open interval ]0,1[ then If has support contained in ]0,1[ iff $(If)(1)=0$.

Main question: Does there exist a ${C}^{\mathrm{\infty}}$ function f with support contained in the open interval ]0,1[ such that ${I}^{k}f$ has support contained in ]0,1[ for every $k\ge 0$, or, equivalently, $({I}^{k}f)(1)=0$ for all $k\ge 0$?

Equivalent formulation: Does there exists a sequence $({f}_{k}{)}_{k\in \mathbb{Z}}$ of ${C}^{\mathrm{\infty}}$ functions each with support contained in the open interval ]0,1[, such that ${f}_{k-1}$ is the derivative of ${f}_{k}$?

Weaker question: Does there at least exist a continuous function f with the properties demanded in the main question?

Stronger question: Does there exist a ${C}^{\mathrm{\infty}}$ function f with compact support, whose Fourier transform vanishes identically on a nontrivial interval?

(A positive answer to the latter would imply a positive answer to the main question: rescale the function so its support is contained in ]0,1[, multiply it appropriately so its Fourier transform vanishes in a neighborhood of 0, and observe that the Fourier transform of ${I}^{k}f$ is, up to constants, ${\xi}^{k}$ times that of f.)

Edit: Before someone points out that the identically zero function fits the bill, I should add that I want my functions to not vanish identically.

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