Using these facts, it is not difficult to evaluate
$\underset{x\to \infty}{\text{lim}}c{x}^{n}$ and
$\underset{x\to \text{\u2212}\infty}{\text{lim}}c{x}^{n},$ where
$c$ is any constant and
$n$ is a positive integer. If
$c>0,$ the graph of
$y=c{x}^{n}$ is a vertical stretch or compression of
$y={x}^{n},$ and therefore
If
$c=0,y=c{x}^{n}=0,$ in which case
$\underset{x\to \infty}{\text{lim}}c{x}^{n}=0=\underset{x\to \text{\u2212}\infty}{\text{lim}}c{x}^{n}.$
Limits at infinity for power functions
For each function
$f,$ evaluate
$\underset{x\to \infty}{\text{lim}}f\left(x\right)$ and
$\underset{x\to \text{\u2212}\infty}{\text{lim}}f\left(x\right).$
$f\left(x\right)=\mathrm{-5}{x}^{3}$
$f\left(x\right)=2{x}^{4}$
Since the coefficient of
${x}^{3}$ is
$\mathrm{-5},$ the graph of
$f\left(x\right)=\mathrm{-5}{x}^{3}$ involves a vertical stretch and reflection of the graph of
$y={x}^{3}$ about the
$x$ -axis. Therefore,
$\underset{x\to \infty}{\text{lim}}\left(\mathrm{-5}{x}^{3}\right)=\text{\u2212}\infty $ and
$\underset{x\to \text{\u2212}\infty}{\text{lim}}\left(\mathrm{-5}{x}^{3}\right)=\infty .$
Since the coefficient of
${x}^{4}$ is
$2,$ the graph of
$f\left(x\right)=2{x}^{4}$ is a vertical stretch of the graph of
$y={x}^{4}.$ Therefore,
$\underset{x\to \infty}{\text{lim}}2{x}^{4}=\infty $ and
$\underset{x\to \text{\u2212}\infty}{\text{lim}}2{x}^{4}=\infty .$
We now look at how the limits at infinity for power functions can be used to determine
$\underset{x\to \text{\xb1}\infty}{\text{lim}}f\left(x\right)$ for any polynomial function
$f.$ Consider a polynomial function
For example, the function
$f\left(x\right)=5{x}^{3}-3{x}^{2}+4$ behaves like
$g\left(x\right)=5{x}^{3}$ as
$x\to \text{\xb1}\infty $ as shown in
[link] and
[link] .
A polynomial’s end behavior is determined by the term with the largest exponent.
$x$
$10$
$100$
$1000$
$f\left(x\right)=5{x}^{3}-3{x}^{2}+4$
$4704$
$\mathrm{4,970,004}$
$\mathrm{4,997,000,004}$
$g\left(x\right)=5{x}^{3}$
$5000$
$\mathrm{5,000,000}$
$\mathrm{5,000,000,000}$
$x$
$\mathrm{-10}$
$\mathrm{-100}$
$\mathrm{-1000}$
$f\left(x\right)=5{x}^{3}-3{x}^{2}+4$
$\mathrm{-5296}$
$\mathrm{-5,029,996}$
$\mathrm{-5,002,999,996}$
$g\left(x\right)=5{x}^{3}$
$\mathrm{-5000}$
$\mathrm{-5,000,000}$
$\mathrm{-5,000,000,000}$
End behavior for algebraic functions
The end behavior for rational functions and functions involving radicals is a little more complicated than for polynomials. In
[link] , we show that the limits at infinity of a rational function
$f\left(x\right)=\frac{p\left(x\right)}{q\left(x\right)}$ depend on the relationship between the degree of the numerator and the degree of the denominator. To evaluate the limits at infinity for a rational function, we divide the numerator and denominator by the highest power of
$x$ appearing in the denominator. This determines which term in the overall expression dominates the behavior of the function at large values of
$x.$
Determining end behavior for rational functions
For each of the following functions, determine the limits as
$x\to \infty $ and
$x\to \text{\u2212}\infty .$ Then, use this information to describe the end behavior of the function.
$f\left(x\right)=\frac{3x-1}{2x+5}$ (Note: The degree of the numerator and the denominator are the same.)
$f\left(x\right)=\frac{3{x}^{2}+2x}{4{x}^{3}-5x+7}$ (Note: The degree of numerator is less than the degree of the denominator.)
$f\left(x\right)=\frac{3{x}^{2}+4x}{x+2}$ (Note: The degree of numerator is greater than the degree of the denominator.)
The highest power of
$x$ in the denominator is
$x.$ Therefore, dividing the numerator and denominator by
$x$ and applying the algebraic limit laws, we see that
Since
$\underset{x\to \text{\xb1}\infty}{\text{lim}}f\left(x\right)=\frac{3}{2},$ we know that
$y=\frac{3}{2}$ is a horizontal asymptote for this function as shown in the following graph.
Since the largest power of
$x$ appearing in the denominator is
${x}^{3},$ divide the numerator and denominator by
${x}^{3}.$ After doing so and applying algebraic limit laws, we obtain
As
$x\to \text{\xb1}\infty ,$ the denominator approaches
$1.$ As
$x\to \infty ,$ the numerator approaches
$+\infty .$ As
$x\to \text{\u2212}\infty ,$ the numerator approaches
$\text{\u2212}\infty .$ Therefore
$\underset{x\to \infty}{\text{lim}}f\left(x\right)=\infty ,$ whereas
$\underset{x\to \text{\u2212}\infty}{\text{lim}}f\left(x\right)=\text{\u2212}\infty $ as shown in the following figure.
If you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)
Faysal
I don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning
Amor
if you want some help on l hospital rule ask me
Jawad
it's spelled hopital
Connor
hi
BERNANDINO
you are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'Hôpital.
Leo
I had no clue this was an online app
Connor
Total online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were $48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach $60 billion?
you have to apply the function arcsin in both sides and you get
arcsin y = acrsin (sin x)
the the function arcsin and function sin cancel each other so the ecuation becomes
arcsin y = x
you can also write
x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x)
y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
You can factorize the numerator of an expression. What's the problem there? here's an example.
f(x)=((x^2)-(y^2))/2
Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y).
so the overall function becomes :
((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
follow algebraic method. look under factoring numerator from Khan academy