Question -
Answer -
Givenf:┬аR┬атЖТ┬аR+┬атЖТ [4, тИЮ) given by┬аf(x) =┬аx2┬а+4.
Now we have to showthat f is invertible,
Consider injectionof┬аf:
Let┬аx┬аand┬аy┬аbe two elements of the domain (Q),
Such that f(x) =f(y)┬а
тЗТ┬аx2┬а+4 = y2┬а+ 4
тЗТ┬аx2┬а=y2
тЗТ┬аx┬а=┬аy┬а┬а ┬а┬а(as┬аco-domain┬аas┬аR+)
So,┬аf┬аisone-one
Now surjectionof┬аf:
Let┬аy┬аbe in the co-domain (Q),
Such that┬аf(x) =y
тЗТ x2┬а+4 = y
тЗТ x2┬а=y тАУ 4
тЗТ x = тИЪ (y-4) in R
тЗТ┬аf┬аis onto.
So,┬аf┬аis a bijection and, hence, it is invertible.
Now we have tofind┬аf-1:
Let┬аfтИТ1┬а(x)┬а=┬аyтАжтАж(1)
тЗТ┬аx┬а=┬аf┬а(y)
тЗТ┬аx┬а=┬аy2┬а+┬а4
тЗТ┬аx┬атИТ┬а4┬а=┬аy2
тЗТ y = тИЪ (x-4)
So, f-1(x)= тИЪ (x-4)
Now substituting thisvalue in (1) we get,
So, f-1(x)= тИЪ (x-4)