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???]