Question -
Answer -
Givenf:┬аR┬атЖТ┬аR┬аis a function defined by┬аf(x) = 4x3┬а+7
Injectivity:
Let┬аx┬аand┬аy┬аbeany two elements in the domain (R), such that┬аf(x) = f(y)
тЗТ┬а4x3┬а+7┬а=┬а4y3┬а+┬а7
тЗТ┬а4x3┬а=┬а4y3
тЗТ┬аx3┬а=┬аy3
тЗТ┬аx┬а=┬аy
So,┬аf┬аisone-one.
Surjectivity:
Let┬аy┬аbe any element in the co-domain┬а(R),┬аsuchthat┬аf(x) = y┬аfor some element┬аx┬аin┬аR (domain)
f(x) = y
тЗТ┬а4x3┬а+7┬а=┬аy
тЗТ┬а4x3┬а=┬аy┬атИТ7
тЗТ x3┬а=(y тАУ 7)/4
тЗТ x = тИЫ(y-7)/4 in R
So, for every elementin the co-domain, there exists some pre-image in the domain. f┬аis onto.
Since,┬аf┬аis both one-to-one and onto, it is a bijection.