How do you determine #h(x)=f(x)g(x)# and #k(x)=f(x)/g(x)# given #f(x)=sqrt(x-1)# and #g(x)=sqrt(6-x)#?

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Answer 1 · 2017-02-22T08:58:09

#h(x)=sqrt(-x^2+7x-6)# and #k(x)=sqrt((x-1)/(6-x))#

As #f(x)=sqrt(x-1)# and #g(x)=sqrt(6-x)#

#h(x)=f(x)g(x)#

= #sqrt(x-1)sqrt(6-x)#

= #sqrt((x-1)(6-x))#

= #sqrt(6x-x^2-6+x)=sqrt(-x^2+7x-6)#

and #k(x)=f(x)/g(x)#

= #sqrt(x-1)/sqrt(6-x)#

= #sqrt((x-1)/(6-x))#