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Real analysis

A. let f(x):= 1/(x^2), x does not equal zero, x elements of Reals
determine the direct image f(E) where E= {x elements Reals: 1<=x<=2}
determine the inverse image f^-1(G) where G= { x element of Reals: 1<=x<=4}

b. show that if a<b then a< 1/2(a+b)<b

c. for a,b,c elements of reals with a<b, find an explicit bijection of A:={x:a<x<b} onto B:={y: 0<y<1}, show how you derived.

d. if a,b elements of Reals show absval(a+b) = abs(a) + abs(b) if and only if ab >= 0

e. Find all x elements of reals that satisfy the following inequalities
i. abs(x-1) > abs(x+1) ii. abs(x) + abs(x+1) < 2

Solution Preview

(a) since f is decreasing f(E) = { y : 1/4 =< y =< 1 };

f^{-1}(G) = {x in R : f(x) in G } = {x in R : 1/2 = < x =< 1}

(b) if ab >= 0, then either (a >= 0 and b >= 0) or (a =< 0 and b =< 0);

case 1: both a, b >= 0; then a + b >= 0, so |a + b| = a + b = |a| + |b|;

case 2 : both a, b =< 0; then a + b =< 0 and |a| = -a and |b| = -b, so |a + b| = -(a + b) = -a + (-b) = |a| + |b|

Conversely, assume |a + b| = |a| + |b|; then on ...

Solution Summary

This provides examples of working with analysis, including bijections, absolute value, and images.

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