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In this section, you will:
• Find the inverse of a polynomial function.
• Restrict the domain to find the inverse of a polynomial function.

A mound of gravel is in the shape of a cone with the height equal to twice the radius.

The volume is found using a formula from elementary geometry.

We have written the volume $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ in terms of the radius $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ However, in some cases, we may start out with the volume and want to find the radius. For example: A customer purchases 100 cubic feet of gravel to construct a cone shape mound with a height twice the radius. What are the radius and height of the new cone? To answer this question, we use the formula

$r=\sqrt[3]{\frac{3V}{2\pi }}$

This function is the inverse of the formula for $\text{\hspace{0.17em}}V\text{\hspace{0.17em}}$ in terms of $\text{\hspace{0.17em}}r.$

In this section, we will explore the inverses of polynomial and rational functions and in particular the radical functions we encounter in the process.

Finding the inverse of a polynomial function

Two functions $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ are inverse functions if for every coordinate pair in there exists a corresponding coordinate pair in the inverse function, $\text{\hspace{0.17em}}g,\left(b,\text{\hspace{0.17em}}a\right).\text{\hspace{0.17em}}$ In other words, the coordinate pairs of the inverse functions have the input and output interchanged.

For a function to have an inverse function    the function to create a new function that is one-to-one and would have an inverse function.

For example, suppose a water runoff collector is built in the shape of a parabolic trough as shown in [link] . We can use the information in the figure to find the surface area of the water in the trough as a function of the depth of the water.

Because it will be helpful to have an equation for the parabolic cross-sectional shape, we will impose a coordinate system at the cross section, with $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ measured horizontally and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ measured vertically, with the origin at the vertex of the parabola. See [link] .

From this we find an equation for the parabolic shape. We placed the origin at the vertex of the parabola, so we know the equation will have form $\text{\hspace{0.17em}}y\left(x\right)=a{x}^{2}.\text{\hspace{0.17em}}$ Our equation will need to pass through the point (6, 18), from which we can solve for the stretch factor $\text{\hspace{0.17em}}a.$

Our parabolic cross section has the equation

$y\left(x\right)=\frac{1}{2}{x}^{2}$

We are interested in the surface area of the water, so we must determine the width at the top of the water as a function of the water depth. For any depth $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ the width will be given by $\text{\hspace{0.17em}}2x,\text{\hspace{0.17em}}$ so we need to solve the equation above for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and find the inverse function. However, notice that the original function is not one-to-one, and indeed, given any output there are two inputs that produce the same output, one positive and one negative.

To find an inverse, we can restrict our original function to a limited domain on which it is one-to-one. In this case, it makes sense to restrict ourselves to positive $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values. On this domain, we can find an inverse by solving for the input variable:

This is not a function as written. We are limiting ourselves to positive $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ values, so we eliminate the negative solution, giving us the inverse function we’re looking for.

can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
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
Abena
find the equation of the line if m=3, and b=-2
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
Where do the rays point?
Spiro
x=-b+_Гb2-(4ac) ______________ 2a
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
so good
abdikarin
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
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
William
what is f(x)=
I don't understand
Joe
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
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
Thomas
Now it shows, go figure?
Thomas