# Math Help - Function related Problem

1. ## Function related Problem

Problem: Let function f:R->R defined by f(x)=x^2+3x+4

(a) Show f is NOT bijective(one-to-one/onto) {I got this one!)

(b) Find all pairs r1, r2 of real numbers such that f(r1)=f(r2)
Solve:
Let f(r1)=f(r2), then r1^2+3r1+4=r2^2+3r2+4, r1^2-r2^2+3r1-3r2=0, (r1-r2)(r1+r2)+3(r1-r2)=0, (r1-r2)(r1+r2+3)=0, so r1=r2 or r1=-r2-3
It's a parabola graph, and symmetric at line x=-3/2, the lowest point of the parabola graph is at (-3/2, 7/4), only at the lowest point (-3/2, 7/4) of the graph r1=r2=-3/2, r1=-r2-3 for all the rest of the points on the graph. Am I explaining it or concluding it correctly?
[I don't know whether how I conclude is correct or not. Need assistant on this one!!! Is there any better way to state it more professionally???]

(c) Find the set S of all real numbers such that if s belongs to S, then there is no real number x such that f(x)=s
Solve: Let f(x)=s and solve for x, then x^2+3x+4=s, so x=(+/-)1/2sqrt(4s-7)-3/2, when s<7/4 there is no real number solution for x,
therfore the set S={s belongs to R: s<7/4} [Is my conclusion correct???]

(d) What well-known set is the set S in (c) related to?
[How to answer this one? I highly doubt my answer is correct!!!]
The set S in (c) is related to the range(f)???

2. Originally Posted by cxc001
Problem: Let function f:R->R defined by f(x)=x^2+3x+4

(a) Show f is NOT bijective(one-to-one/onto) {I got this one!)

(b) Find all pairs r1, r2 of real numbers such that f(r1)=f(r2)
Solve:
Let f(r1)=f(r2), then r1^2+3r1+4=r2^2+3r2+4, r1^2-r2^2+3r1-3r2=0, (r1-r2)(r1+r2)+3(r1-r2)=0, (r1-r2)(r1+r2+3)=0, so r1=r2 or r1=-r2-3
It's a parabola graph, and symmetric at line x=-3/2, the lowest point of the parabola graph is at (-3/2, 7/4), only at the lowest point (-3/2, 7/4) of the graph r1=r2=-3/2, r1=-r2-3 for all the rest of the points on the graph. Am I explaining it or concluding it correctly?
[FONT=&quot][SIZE=3] [I don't know whether how I conclude is correct or not. Need assistant on this one!!! Is there any better way to state it more professionally???]
The layout of your answer is cramped and difficult to follow, improve it and you will get more credit for what you have done correctly as the marker will be able to see what you are trying to say.

You are asked to find all pairs of reals $r_1$ and $r_2$, you should end by saying something like: all pairs of non-equal reals $x_1,\ x_2$ for which $f(x_1)=f(x_2)$ are of the form: $r_1= ...$ and $r_2= ...$

CB

3. Originally Posted by cxc001
Problem: Let function f:R->R defined by f(x)=x^2+3x+4

(a) Show f is NOT bijective(one-to-one/onto) {I got this one!)

(b) Find all pairs r1, r2 of real numbers such that f(r1)=f(r2)
Solve:
Let f(r1)=f(r2), then r1^2+3r1+4=r2^2+3r2+4, r1^2-r2^2+3r1-3r2=0, (r1-r2)(r1+r2)+3(r1-r2)=0, (r1-r2)(r1+r2+3)=0, so r1=r2 or r1=-r2-3
It's a parabola graph, and symmetric at line x=-3/2, the lowest point of the parabola graph is at (-3/2, 7/4), only at the lowest point (-3/2, 7/4) of the graph r1=r2=-3/2, r1=-r2-3 for all the rest of the points on the graph. Am I explaining it or concluding it correctly?
[I don't know whether how I conclude is correct or not. Need assistant on this one!!! Is there any better way to state it more professionally???]

(c) Find the set S of all real numbers such that if s belongs to S, then there is no real number x such that f(x)=s
Solve: Let f(x)=s and solve for x, then x^2+3x+4=s, so x=(+/-)1/2sqrt(4s-7)-3/2, when s<7/4 there is no real number solution for x,
therfore the set S={s belongs to R: s<7/4} [Is my conclusion correct???]
Part (c) is correct (given the vagaries of having to type it in plain text), but once again layout needs improving.

(Part (d) is to vague a question for me to help you with, there is probably something in your notes that your teacher wants you to refer to, possibly the negative reals?)

CB