Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 9E from Chapter 1.1 from Stewart's Calculus, 8th Edition.
Problem 9E
Chapter:
Problem:
Determine whether the curve is the graph of a function of x. If it is, state the domain and range of the function.
Step-by-Step Solution
Step 1
We are given with following graph
https://imgur.com/3vQah6T
Step 2:
As we can see that if we draw any vertical line, it never intersects the graph at more than one point.
So the given graph represents a function Therefore,
It is a Function
Step 3: Domain
Domain of the function is all possible values of x for which graph exists (or function value are possible)
As we can see that graph exists for all x value between x=-3 and x=2 including boundaries. So domain is:Therefore,
\[{D_f} = \left[ { - 3,2} \right]\]
Step 4: Range
Range of the function is all possible values of y or function value that can be found on the graph
As we can see that graph starts at y=-3 and goes up to y=3 (both inclusive). However the graph does not exist for y values between -2 and -1. So range of the function is: \[{R_f} = \left[ { - 3,3} \right] - [ - 2, - 1) = [ - 3, - 2)U[ - 1,3]\]Therefore,
\[{R_f} = [ - 3, - 2)U[ - 1,3]\]
We are given with following graph
https://imgur.com/3vQah6T
Step 2:
As we can see that if we draw any vertical line, it never intersects the graph at more than one point.
So the given graph represents a function Therefore,
It is a Function
Step 3: Domain
Domain of the function is all possible values of x for which graph exists (or function value are possible)
As we can see that graph exists for all x value between x=-3 and x=2 including boundaries. So domain is:Therefore,
\[{D_f} = \left[ { - 3,2} \right]\]
Step 4: Range
Range of the function is all possible values of y or function value that can be found on the graph
As we can see that graph starts at y=-3 and goes up to y=3 (both inclusive). However the graph does not exist for y values between -2 and -1. So range of the function is: \[{R_f} = \left[ { - 3,3} \right] - [ - 2, - 1) = [ - 3, - 2)U[ - 1,3]\]Therefore,
\[{R_f} = [ - 3, - 2)U[ - 1,3]\]