What is the Domain and Range of a Function with Given Constraints?

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Homework Help Overview

The discussion revolves around determining the domain and range of functions, specifically focusing on a quadratic function f(x) = x^2 + 3 with a restricted domain of -3 ≤ x ≤ 3. Participants are exploring algebraic methods to find the range and discussing the implications of the function's properties.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to find the range of the quadratic function algebraically, questioning the validity of their steps and the assumptions made about inequalities. There is also a discussion about the nature of the function and its invertibility, as well as the implications of the domain given.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. Some guidance has been offered regarding the assumptions needed for manipulating inequalities, and there is recognition of the need to consider the restricted domain in the context of the function's behavior.

Contextual Notes

Participants are navigating constraints related to the function's domain and the properties of quadratic functions. There is mention of the need for additional assumptions when manipulating inequalities, and some participants express confusion regarding the relationship between the functions discussed.

preet
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I'm a little confused here. I have a function: f(x) = x^2 + 3
If the domain of the function is

-3 <= x <= 3

I need to find the range...

I think that the answer is 3 <= y <= 12... that can be found out by drawing the graph. But I want to find the range algebraically. It says in my textbook to turn the 'x' in the middle of the domain shown above into the respective f(x) equation (x^2)+3... and performing whatever action on x on the other sides of the inequality. This works in other functions, but I think I'm doing something wrong with this one:

1.-3 <= x <= 3
2.9 <= x^2 <= 9
3.12 <= x^2 + 3 <= 12

I think that because in step 1 I multiplied one value by (-3) I need to flip the greater than/equal to sign(s), but not sure how to do it... as you can see, by step 3, the wrong answer is reached. Can anyone show me my mistake? Thanks!
 
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Hi preet - just curious...
how did you come up with that domain? what's the problem with x = -4 or x = 4?x = -5 or x = 5? etc?
 
The method you used only works if f(x) has a true inverse function which in this case it doesn't. This is because more than one x value exists for every y which is untrue of truly invertible functions. The function f(x) where x is lies on [0, \infty) does have a true inverse. Take the domain to be 0<=x<=3 and then you can apply the rules you used.
 
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ah, nevermind.. preet must be referring to a restricted domain.
 
To go from an inequality like a<b to a^2<b^2 you need the additional assumption that a and b are non-negative 0<=a<b. Otherwise it may not be true, -4<3 but (-4)^2>3^2.

You can go a couple ways, do 2 cases, one where x<=0 and one where x>0. Or you can recognize that -3<=x<=3 is equivalent to |x|<=3, and work with 0<=|x|<=3.
 
Last edited:
The domain was given... anyways... I understand now. Thanks!
 
I have another question related to functions... I thought I'd post it in here instead of starting a new thread...

A graph of y=g(x) is as follows:
(-4, 3), (-1, -2), (2, -2), (3, 1), (6, 1)

the graph doesn't show a relation... it just shows a function... and the graph isn't a plot, so it isn't discrete.

f(x) = -g (1/2x - 3) + 1

Anyways, the questions are:
a) Draw an input/output diagram for f.
b) Determine the domain and range of f.
c) Graph 'f'

All I can do so far is
a) [ add 6 ] + [ multiply by 2 ] + [ g ] + [ multiply by -1 ] + [ add 2 ]
b) I don't know how to figure this out... the fact that function g is in function f is really confusing me...
c) I don't know how to do this either... I was thinking that I have to apply the transformations in the input output diagram, but I want to figure out part b first.

TiA.
 

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