Restrict Domain of Function...3

[nycmathguy]

Section 1.9
Question 76

Had to go back to Section 1.9 for one more question concerning the restriction of the domain of a function to pass the horizontal line test.
Take a look at question 76.

The function y = (1/2)x^2 - 1 fails the horizontal line test, which means it has no inverse. I must restrict its domain. My struggle is forming a function after restricting the domain of the original problem.

See picture of graph. I have decided to cut the left side of the parabola. When I do, what does the original function become? After cutting the left side of the parabola, I am left with the right side only from the point (0, -1) to infinity.

Stuck here.

 

[MathLover1]

y = (1/2)x^2 - 1 is a parabola with a vertex at (0,-1), means y-axis is axis of symmetry

if you restrict a domain to x>=0, you get right half of parabola and it will pass horizontal test

or
if you restrict a domain to x<=0, you get left half of parabola and it will pass horizontal test

 

[nycmathguy]

y = (1/2)x^2 - 1 is a parabola with a vertex at (0,-1), means y-axis is axis of symmetry

if you restrict a domain to x>=0, you get right half of parabola and it will pass horizontal test

View attachment 282


or
if you restrict a domain to x<=0, you get left half of parabola and it will pass horizontal test
View attachment 283

I understand but what is the function for half of a parabola? In other words, y = what???

 

[MathLover1]

this is your parabola

here, right half is brawn

this is a part of y = (1/2)x^2 - 1 with restricted domain; x>=0

so, you just making a table using values x>=0 , plot points and draw a curve through

 

[nycmathguy]

this is your parabola
View attachment 286

here, right half is brawn
View attachment 285


this is a part of y = (1/2)x^2 - 1 with restricted domain; x>=0

View attachment 288

so, you just making a table using values x>=0 , plot points and draw a curve through

Sorry but this remains unclear to me. Please, finish this problem in step by step fashion. Can you also completely do 78? This is one of those problems that I just can't quite connect with. Know what I mean?

 

[MathLover1]

y = (1/2)x^2 - 1

imagine you need to draw this parabola
you need table

x| y using only positive values of x

1| -1/2.................y = (1/2)*1^2 - 1=-1/2
2| 1 .................y = (1/2)*2^2 - 1=1
3| 7/2 .................y = (1/2)*3^2 - 1=7/2
4| 7 .................y = (1/2)*4^2 - 1=7

plot these points

and draw a line through

you can also choose left side, do same table using x =,0,-1,-2,-3,-4

 

[nycmathguy]

y = (1/2)x^2 - 1

imagine you need to draw this parabola
you need table

x| y using only positive values of x

1| -1/2.................y = (1/2)*1^2 - 1=-1/2
2| 1 .................y = (1/2)*2^2 - 1=1
3| 7/2 .................y = (1/2)*3^2 - 1=7/2
4| 7 .................y = (1/2)*4^2 - 1=7

plot these points
View attachment 289


and draw a line through

View attachment 290


you can also choose left side, do same table using x =,0,-1,-2,-3,-4

But did you answer the question as stated in the textbook?

 

[MathLover1]

1. The function y = (1/2)x^2 - 1 fails the horizontal line test, which means it has no inverse. Then, only option is to restrict domain the way I did it above.

here is one link for you to see similar case

 

[nycmathguy]

1. The function y = (1/2)x^2 - 1 fails the horizontal line test, which means it has no inverse. Then, only option is to restrict domain the way I did it above.

here is one link for you to see similar case

I will check it out. Notice what the set of instructions say to do:

•restrict the domain
•find the inverse
•state the domain and range of f
•state the domain and range of f inverse