Determine where the function
$\text{\hspace{0.17em}}f(x)=\{\begin{array}{l}\frac{\pi x}{4},\text{}x2\\ \frac{\pi}{x},\text{}2\le x\le 6\\ 2\pi x,\text{}x6\end{array}\text{\hspace{0.17em}}$ is discontinuous.
To determine whether a
piecewise function is continuous or discontinuous, in addition to checking the boundary points, we must also check whether each of the functions that make up the piecewise function is continuous.
Given a piecewise function, determine whether it is continuous.
Determine whether each component function of the piecewise function is continuous. If there are discontinuities, do they occur within the domain where that component function is applied?
For each boundary point
$\text{\hspace{0.17em}}x=a\text{\hspace{0.17em}}$ of the piecewise function, determine if each of the three conditions hold.
Determining whether a piecewise function is continuous
Determine whether the function below is continuous. If it is not, state the location and type of each discontinuity.
The two functions composing this piecewise function are
$\text{\hspace{0.17em}}f(x)=\mathrm{sin}(x)\text{\hspace{0.17em}}$ on
$\text{\hspace{0.17em}}x<0\text{\hspace{0.17em}}$ and
$\text{\hspace{0.17em}}f(x)={x}^{3}\text{\hspace{0.17em}}$ on
$\text{\hspace{0.17em}}x>0.\text{\hspace{0.17em}}$ The sine function and all polynomial functions are continuous everywhere. Any discontinuities would be at the boundary point,
At
$\text{\hspace{0.17em}}x=0,$ let us check the three conditions of continuity.
Because all three conditions are not satisfied at
$\text{\hspace{0.17em}}x=0,$ the function
$\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ is discontinuous at
$\text{\hspace{0.17em}}x=0.$
A function has a jump discontinuity if the left- and right-hand limits are different, causing the graph to “jump.”
A function has a removable discontinuity if it can be redefined at its discontinuous point to make it continuous. See
[link] .
Some functions, such as polynomial functions, are continuous everywhere. Other functions, such as logarithmic functions, are continuous on their domain. See
[link] and
[link] .
For a piecewise function to be continuous each piece must be continuous on its part of the domain and the function as a whole must be continuous at the boundaries. See
[link] and
[link] .
Section exercises
Verbal
State in your own words what it means for a function
$\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ to be continuous at
$\text{\hspace{0.17em}}x=c.$
Informally, if a function is continuous at
$\text{\hspace{0.17em}}x=c,$ then there is no break in the graph of the function at
$\text{\hspace{0.17em}}f\left(c\right),$ and
$\text{\hspace{0.17em}}f\left(c\right)\text{\hspace{0.17em}}$ is defined.
For the following exercises, determine why the function
$\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is discontinuous at a given point
$\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ on the graph. State which condition fails.
Discontinuous at
$\text{\hspace{0.17em}}a=3$ ;
$\text{\hspace{0.17em}}\underset{x\to 3}{\mathrm{lim}}f(x)=3,$ but
$\text{\hspace{0.17em}}f(3)=6,$ which is not equal to the limit.
Someone should please solve it for me
Add 2over ×+3 +y-4 over 5
simplify (×+a)with square root of two -×root 2 all over a
multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15
Second one, I got Root 2
Third one, I got 1/(y to the fourth power)
I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
graph the following linear equation using intercepts method.
2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b
you were already given the 'm' and 'b'.
so..
y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line.
where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2
2=3x
x=3/2
then .
y=3/2X-2
I think
Given
co ordinates for x
x=0,(-2,0)
x=1,(1,1)
x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
I've run into this:
x = r*cos(angle1 + angle2)
Which expands to:
x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2))
The r value confuses me here, because distributing it makes:
(r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1))
How does this make sense? Why does the r distribute once
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
Brad
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis
vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As
'f(x)=y'.
According to Google,
"The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
GREAT ANSWER THOUGH!!!
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks.
"Â" or 'Â' ... Â