Expert: Paul Klarreich - 5/15/2007

Question
hi i am trying to complete this question. the diagram is one of a graph of the function f(x)=(x^2)+1,-1<=x<=1 and p,q E(element) R.(i)find value of p & q.[coordinates are at (-1,p) & (q,2). Ans-: I found p to be '2' and q to be '1' by subbing directly into the function (ii)the range of the function f(x) for the given domain Ans-: -1<=x<=1 (I got this range directly from the given info; is there a way to show working?)(b(i)) Determine whether f(x) is surjective(onto) (ii)is injective (one-to-one) (iii) has an inverse.
Now, I am not sure how to do these last ones. could you please help me. thank you in advance for you help

Answer
Questioner:   jon
Category:  Calculus
Private:  No
 
Subject:  math problem
Question:  hi i am trying to complete this question. the diagram is one of a graph of the function f(x)=(x^2)+1,-1<=x<=1 and p,q E(element) R.(i)find value of p & q.[coordinates are at (-1,p) & (q,2). Ans-: I found p to be '2' and q to be '1' by subbing directly into the function (ii)the range of the function f(x) for the given domain Ans-: -1<=x<=1 (I got this range directly from the given info; is there a way to show working?)(b(i)) Determine whether f(x) is surjective(onto) (ii)is injective (one-to-one) (iii) has an inverse.
Now, I am not sure how to do these last ones. could you please help me. thank you in advance for you help
.......................
Hi, Jon,

For your function:  f(x) = x^2 + 1, on [-1,1].

If the graph of f passes through  (-1,p) and (q,2)

Yes, p = 2 looks right.  However, I think you should say  q = +- 2.

The range of f is certainly  [1,2].  Why?  I think you can prove the range to be [1,2] in this way:

f(x) = x^2 + 1  is continuous, because it is a polynomial.  

The max and min values on [-1,1] can be found by the usual calculus work of finding critical points and testing.  OR, you just note that the graph is a parabola and its possible extreme points are the turning point, x = 0, and the endpoints of your interval.

f(0) = 1
f(+-1) = 2

So we find the max is y = 2, and the min is y = 1.  Since f(x) is continuous, it takes on all values between those and the range is [1,2].

Now about your function f as a mapping:  [-1,1] --> [1,2]:

Since f takes on all values in [1,2] it is ONTO.

Since f(-x) = f(x) for all such x, it is NOT 1-1.  As such, it does not have an inverse.