Question

Sa se arate ca urmatoarele functii sunt injective:
a)f:R=>R , f(x)=2x+5
b)f:R=>R, f(x)=-5x+2
c)f:[0,infinit)=>[3,infinit) , f(x)=2x+3
d)f:R-{1}=>R, f(x)=1-x supra 1+x
Dau coroana!

Answer 1

$\displaystyle a,~b~si~c)~La~primele~3~puncte~avem~cate~o~functie~de~gradul~1. \\ \\ Insa~orice~functie~de~gradul~1~este~injectiva.~Demonstratie: \\ \\ Fie~functia~f: A\rightarrow A,~f(x)=ax+b~cu~a,b \in \mathbb{R},~a \neq 0. \\ \\ Functia~este~injectiva~daca~si~numai~daca~f(x)=f(y)~implica \\ \\ x=y. ~(unde~x,y \in A~;~A~fiind~o~multime~nevida~oarecare)\\ \\ f(x)=f(y) \Leftrightarrow ax+b=ay+b \Leftrightarrow ax-ay+b-b=0 \Leftrightarrow \\ \\ \Leftrightarrow a(x-y)=0 \Leftrightarrow x=y.$

$\displaystyle Deci~f(x)=f(y)~implica~x=y,~ceea~ce~inseamna~ca~f~este~ \\ \\ injectiva. \\ \\ d)~Rezolvam~ecuatia~f(x)=f(y),~cu~x,y \in \mathbb{R}- \{-1\}. \\ \\ f(x)=f(y) \Leftrightarrow \frac{1-x}{1+x}= \frac{1-y}{1+y} \Leftrightarrow (1-x)(1+y)=(1-y)(1+x) \Leftrightarrow \\ \\ \Leftrightarrow 1+y-x-xy=1+x-y-xy \Leftrightarrow y-x=x-y \Leftrightarrow 2y=2x \Leftrightarrow \\ \\ \Leftrightarrow x=y. \\ \\ Deci~f(x)=f(y)~implica~x=y,~ceea~ce~inseamna~ca~functia~ \\ \\ este~injectiva.$