Inverse function of a quadratic and a cubic function?

AI Thread Summary
The discussion revolves around finding the inverse functions of a quadratic and a cubic function. For the quadratic function, the inverse results in a negative value under a square root, leading to complex numbers, which complicates the calculation of y-values. The user expresses frustration with their calculator errors and seeks clarification on how to graph the inverse function. The cubic function's inverse is given as y=3√x, and the user requests sample calculations for specific x-values to aid in understanding. The conversation highlights the importance of recognizing when a graph does not represent a function and the need for clarity in solving these types of problems.
supernova1203
Messages
209
Reaction score
0

Homework Statement


jruv4i.jpg



Homework Equations





The Attempt at a Solution



for 24. which is a quadratic function, i understand that you have to exchange the positions of x and y and solve from there...but my question is the inverse function of that quadratic for 24 is a negative number under a square root...iv tried google etc... apparently I am dealing with imaginary or complex numbers or something... my calculator keeps giving me an error whenever i try to come up with points for the table of values using the inverse function

the inverse function btw is y=±√x-3

how do i get around this? the book clearly shows that you can graph this inverse function but i can't get the y values because the negative number in square root

for 25. the inverse function is y=3√x

using the x values in table of values of -3, -2, -1, 0 ,1 ,2 ,3

can someone please do a few sample calculations and show me how to get the answer? or how they got the solutions for a few of those points? I am completely lost for question 25.

thanks
 
Physics news on Phys.org
supernova1203 said:

Homework Statement


jruv4i.jpg



Homework Equations





The Attempt at a Solution



for 24. which is a quadratic function, i understand that you have to exchange the positions of x and y and solve from there...but my question is the inverse function of that quadratic for 24 is a negative number under a square root...iv tried google etc... apparently I am dealing with imaginary or complex numbers or something... my calculator keeps giving me an error whenever i try to come up with points for the table of values using the inverse function

the inverse function btw is y=±√x-3

how do i get around this? the book clearly shows that you can graph this inverse function but i can't get the y values because the negative number in square root

for 25. the inverse function is y=3√x

using the x values in table of values of -3, -2, -1, 0 ,1 ,2 ,3

can someone please do a few sample calculations and show me how to get the answer? or how they got the solutions for a few of those points? I am completely lost for question 25.

thanks

In 24, you are missing the point completely. Go back to basics: do you know what a function is (an exact definition, please, not some hand-waving statement). Do you know what would be meant by an inverse function (if it existed)? Can you see the trouble in example 24? Hint: it has nothing to do with the fact that for some values of y the equation y = 3 + x^2 has no real solutions; obviously, 3 + x^3 ≥ 3 always, so if we pick a value of y that is < 3 there will be no solution. Again, I emphasize: that is NOT the issue. Go back and answer---very carefully--- the questions I have asked you.

RGV
 
ok i was able to solve 24 on my own..still no news on 25 though
 
Your book should have some discussion about how to determine whether a graph represents a function or not. For example, the graph of y2 = x (or equivalently, y = ±√x) does not represent a function.
 
yeah, its simple the vertical line test, what i am trying to figure out is how to get the points so i can graph the inverse function for question 25. The book says the inverse function is y=3√x
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top