How do you prove that the function $f\left(x\right)={x}^{2}-3x+5$ is continuous at a =2?

Zachariah Norris
2022-09-15
Answered

How do you prove that the function $f\left(x\right)={x}^{2}-3x+5$ is continuous at a =2?

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ti1k1le2l

Answered 2022-09-16
Author has **6** answers

Any polynomial function is continuous everywhere, but let's prove this particular example using the limit definition of continuity...

A function f(x) is continuous at a point a if both f(a) and $\underset{x\to a}{lim}f\left(x\right)$ are defined and equal.

In our example:

$f\left(2\right)={2}^{2}-3\left(2\right)+5=4-6+5=3$

$={2}^{2}+4h+{h}^{2}-3\left(2\right)-3h+5$

$={2}^{2}+4h+{h}^{2}-3\left(2\right)-3h+5$

$=4-6+5+h+{h}^{2}$

$=3+h+{h}^{2}\to 3$ as $h\to 0$

So $\underset{x\to 2}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(2+h)=3=f\left(2\right)$

A function f(x) is continuous at a point a if both f(a) and $\underset{x\to a}{lim}f\left(x\right)$ are defined and equal.

In our example:

$f\left(2\right)={2}^{2}-3\left(2\right)+5=4-6+5=3$

$={2}^{2}+4h+{h}^{2}-3\left(2\right)-3h+5$

$={2}^{2}+4h+{h}^{2}-3\left(2\right)-3h+5$

$=4-6+5+h+{h}^{2}$

$=3+h+{h}^{2}\to 3$ as $h\to 0$

So $\underset{x\to 2}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(2+h)=3=f\left(2\right)$

asked 2022-05-10

As part of a larger question regarding proving a subset is not connected, so I have $S=\{(x,y)\in {\mathbb{R}}^{2}:xy=1\}$ and I have said this is not connected as the function below:

$f(x)=\{\begin{array}{ll}1& x>0\\ 0& x<0\end{array}$

is a surjective, continuous function that maps $f:S\to \{0,1\}$

So I now want to show $f(x)$ is continuous, I have attempted to use $li{m}_{x\to a}f(x)=f(a)$ but I can't work out how to get this written out and what values of $x$ I need to take? I'm also doubting whether this is continuous but for all values of $x$ would not include 0 as it cannot take this value?

$f(x)=\{\begin{array}{ll}1& x>0\\ 0& x<0\end{array}$

is a surjective, continuous function that maps $f:S\to \{0,1\}$

So I now want to show $f(x)$ is continuous, I have attempted to use $li{m}_{x\to a}f(x)=f(a)$ but I can't work out how to get this written out and what values of $x$ I need to take? I'm also doubting whether this is continuous but for all values of $x$ would not include 0 as it cannot take this value?

asked 2022-08-25

How do you prove that $g\left(x\right)=\frac{1}{2}x$ is continuous at $x=\frac{1}{4}$?

asked 2022-08-12

If $f(x)=\frac{1}{(x-2)(2x-5)}$ and $g(x)=\frac{1}{{x}^{2}}$ , we define f(g(x)) over the domain in which its defined , is it correct to say f(g(x)) is discontinuous at $+-\surd (2/5)$ ,0 and +-$1/\surd 2$ ? 0 because the input is going into g(x) hence as its not in domain of g(x) so it can be said its discontinuous at that point . and similarily other four points as its not in domain of f(g(x)) ? But if suppose we say h(x)=f(g(x)) will we say h(x) is discontinuous at those points (maybe different ones) where x is not defined after simplyfing the whole f(g(x)) to a rational function ?

asked 2022-06-29

Let $\mathrm{\Omega}=\mathbb{C}\mathrm{\setminus}[-1,1]$, i.e. deleting ``the line'' only, is there a function $f:\mathrm{\Omega}\to \mathbb{C}$ such that $f$ satisfies $f(z{)}^{2}=1-{z}^{2}$ and is continuous on this region?

My guess is that such a function would exist, but requires a piecewise definition. A candidate solution I have been working on is $f(z)={e}^{\frac{1}{2}\mathrm{log}(1-{z}^{2})}$. The problem with this solution is having the domain, as I realize it is possible to have $1-{z}^{2}>1$, where my solution is well-defined. Is there a way to work around this or should I try something else?

My guess is that such a function would exist, but requires a piecewise definition. A candidate solution I have been working on is $f(z)={e}^{\frac{1}{2}\mathrm{log}(1-{z}^{2})}$. The problem with this solution is having the domain, as I realize it is possible to have $1-{z}^{2}>1$, where my solution is well-defined. Is there a way to work around this or should I try something else?

asked 2022-07-02

In real functions, do we have a notion of one-sided measure theoretic limits? I want to define them with the following:

$\underset{x\to {c}^{+}}{lim}f(x)=L$

iff

$\mathrm{\forall}\u03f5,\mathrm{\exists}\delta >0,J:=f((c,c+\delta )),\mu (J\cap {B}_{\u03f5}(L))=\mu (J)$

$\underset{x\to {c}^{+}}{lim}f(x)=L$

iff

$\mathrm{\forall}\u03f5,\mathrm{\exists}\delta >0,J:=f((c,c+\delta )),\mu (J\cap {B}_{\u03f5}(L))=\mu (J)$

asked 2022-09-23

What makes a function continuous at a point?

asked 2022-09-18

Let

$f(x)=\{\begin{array}{lcc}\frac{{z}^{3}-1}{{z}^{2}+z+1}& if& |z|\ne 1\\ \\ \frac{-1+i\sqrt{3}}{2}& if& |z|=1\end{array}$

is $f$ continous in ${z}_{0}=\frac{1+\sqrt{3}i}{2}$

I think to $f$ is not continous at ${z}_{0}$, i try using the sequence criterion for continuity searching a sequence $\{{z}_{n}\}$ such that $\underset{n\to \mathrm{\infty}}{lim}{z}_{n}={z}_{0}$ but $\underset{n\to \mathrm{\infty}}{lim}f({z}_{n})\ne f({z}_{0})$. but i cant find that sequence, ill be very grateful for any hint or help to solve my problem.

$f(x)=\{\begin{array}{lcc}\frac{{z}^{3}-1}{{z}^{2}+z+1}& if& |z|\ne 1\\ \\ \frac{-1+i\sqrt{3}}{2}& if& |z|=1\end{array}$

is $f$ continous in ${z}_{0}=\frac{1+\sqrt{3}i}{2}$

I think to $f$ is not continous at ${z}_{0}$, i try using the sequence criterion for continuity searching a sequence $\{{z}_{n}\}$ such that $\underset{n\to \mathrm{\infty}}{lim}{z}_{n}={z}_{0}$ but $\underset{n\to \mathrm{\infty}}{lim}f({z}_{n})\ne f({z}_{0})$. but i cant find that sequence, ill be very grateful for any hint or help to solve my problem.