gaby131o
2022-09-18
Answered

How do you graph $f\left(x\right)=\frac{3}{x}+1$ using holes, vertical and horizontal asymptotes, x and y intercepts?

You can still ask an expert for help

Wischarm1q

Answered 2022-09-19
Author has **5** answers

y axis intercepts occur when x=0. x=0 is undefined for this function, so no y intercept.

x axis intercepts occur when y=0:

$\frac{3}{x}+1=0\Rightarrow x=-3$

Coordinate: (−3,0)

as $x\to \infty$, ${888}\frac{3}{x}+1\to 1$

as $x\to -\infty$ , ${888}\frac{3}{x}+1\to 1$

The line y=1 is a horizontal asymptote.

as $x\to {0}^{+}$ , ${888}\frac{3}{x}+1\to \infty$

as $x\to {0}^{-}$, ${888}\frac{3}{x}+1\to -\infty$

The line x=0 is a vertical asymptote.

x axis intercepts occur when y=0:

$\frac{3}{x}+1=0\Rightarrow x=-3$

Coordinate: (−3,0)

as $x\to \infty$, ${888}\frac{3}{x}+1\to 1$

as $x\to -\infty$ , ${888}\frac{3}{x}+1\to 1$

The line y=1 is a horizontal asymptote.

as $x\to {0}^{+}$ , ${888}\frac{3}{x}+1\to \infty$

as $x\to {0}^{-}$, ${888}\frac{3}{x}+1\to -\infty$

The line x=0 is a vertical asymptote.

asked 2021-02-25

True or False. The graph of a rational function may intersect a horizontal asymptote.

asked 2022-09-10

What are the excluded values for the rational expression $\frac{8}{{x}^{2}-16}$?

asked 2022-06-25

The form of the partial fraction decomposition of a rational function is given below.

$\frac{x-3{x}^{2}-26}{(x+1)({x}^{2}+9)}=\frac{A}{x+1}+\frac{Bx+C}{{x}^{2}+9}$

What are the values of $A,B$ and $C$?

So can someone explain how to get the answer. I'm not sure how I got to the answer but I know $B=0$ and $C=1$

I also found the indefinite integral, which equals

$\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{x}{3}\right)-3\mathrm{log}(1+x)+C$

$\frac{x-3{x}^{2}-26}{(x+1)({x}^{2}+9)}=\frac{A}{x+1}+\frac{Bx+C}{{x}^{2}+9}$

What are the values of $A,B$ and $C$?

So can someone explain how to get the answer. I'm not sure how I got to the answer but I know $B=0$ and $C=1$

I also found the indefinite integral, which equals

$\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{x}{3}\right)-3\mathrm{log}(1+x)+C$

asked 2022-07-08

Let $f(z)$ be a rational function in the complex plane such that $f$ does not have any poles in $\{z:\mathrm{\Im}z\ge 0\}$.

Prove that $sup\{|f(z)|:\mathrm{\Im}z\ge 0\}=sup\{|f(z)|:\mathrm{\Im}z=0\}$.

Let ${\mathrm{\Gamma}}_{r}$ be a half circle counter such that ${\mathrm{\Gamma}}_{r}={\mathrm{\Gamma}}_{{r}_{1}}+{\mathrm{\Gamma}}_{{r}_{2}}$ when ${\mathrm{\Gamma}}_{{r}_{1}}=\{z:\mathrm{\Im}z=0,|z|=r\}$, ${\mathrm{\Gamma}}_{{r}_{2}}=\{z:\mathrm{\Im}z>0,|z|=r\}$. Using the Maximum modulus principle on the insides of ${\mathrm{\Gamma}}_{r}$, $|f|$ Gets is maximum value on ${\mathrm{\Gamma}}_{r}$. As r gets bigger if |f| got it's maximum on ${\mathrm{\Gamma}}_{{r}_{2}}$ than it's still smaller the the value on ${\mathrm{\Gamma}}_{r+1}$ which does not contain ${\mathrm{\Gamma}}_{{r}_{2}}$. I would like a hint on how to proceed.

Prove that $sup\{|f(z)|:\mathrm{\Im}z\ge 0\}=sup\{|f(z)|:\mathrm{\Im}z=0\}$.

Let ${\mathrm{\Gamma}}_{r}$ be a half circle counter such that ${\mathrm{\Gamma}}_{r}={\mathrm{\Gamma}}_{{r}_{1}}+{\mathrm{\Gamma}}_{{r}_{2}}$ when ${\mathrm{\Gamma}}_{{r}_{1}}=\{z:\mathrm{\Im}z=0,|z|=r\}$, ${\mathrm{\Gamma}}_{{r}_{2}}=\{z:\mathrm{\Im}z>0,|z|=r\}$. Using the Maximum modulus principle on the insides of ${\mathrm{\Gamma}}_{r}$, $|f|$ Gets is maximum value on ${\mathrm{\Gamma}}_{r}$. As r gets bigger if |f| got it's maximum on ${\mathrm{\Gamma}}_{{r}_{2}}$ than it's still smaller the the value on ${\mathrm{\Gamma}}_{r+1}$ which does not contain ${\mathrm{\Gamma}}_{{r}_{2}}$. I would like a hint on how to proceed.

asked 2022-06-15

Using this rule: Integral Calculus "Rational Function Rule"

$\int \frac{1}{{a}^{2}-{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{2a}{\mathrm{log}}_{e}\left(\frac{a+x}{a-x}\right)+c$

I have integrated a function, but my question is if I'm solving for $x$, in the next line I must raise both sides to the base of $e$, and there is still $+c$ (constant) on the end of the integrated function, so do I leave it as plus $c$ on the RHS on the equation (not do anything to it, so when I want to work it out by substituting $({x}_{1},{y}_{1})$ at the end it will be ${x}_{1}=\dots {y}_{1}\dots +c$) or when I raise both sides to the base of $e$ does $c$, the constant become $c\ast e\dots $ instead of $+c$?

Any help would be great.

$\int \frac{1}{{a}^{2}-{x}^{2}}\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{2a}{\mathrm{log}}_{e}\left(\frac{a+x}{a-x}\right)+c$

I have integrated a function, but my question is if I'm solving for $x$, in the next line I must raise both sides to the base of $e$, and there is still $+c$ (constant) on the end of the integrated function, so do I leave it as plus $c$ on the RHS on the equation (not do anything to it, so when I want to work it out by substituting $({x}_{1},{y}_{1})$ at the end it will be ${x}_{1}=\dots {y}_{1}\dots +c$) or when I raise both sides to the base of $e$ does $c$, the constant become $c\ast e\dots $ instead of $+c$?

Any help would be great.

asked 2022-06-21

given the rational function $\frac{1}{{x}^{2}-\frac{x}{2}-3}$ and asked for the domain and range, I multiplied thru by 2 and got $\frac{2}{2{x}^{2}-x-6}$ I understand the domain includes all real numbers except for (vertical asymptotes) $\frac{-3}{2}$ and 2. i thought there were no horizontal asymptotes until I graphed it on a calculator. according to the calculator the range is $(-\mathrm{\infty},\frac{-16}{49}]\cup (0,\mathrm{\infty})$. this makes sense to me looking at the graph but how can I obtain these results for the range using algebra? thank you

asked 2022-02-17

I'm searching the antiderivative of rational functions:

1)

For this one we have

Same thing for 2)

What are the methods to find antiderivative of rational functions like these ones?

Thank you For (1): Try the substitution

Then your integral becomes