How do we Find the maximum and minimum vlaues of 1/(3sinx-4cosx+7)?

1 Answer
Mar 20, 2018

Let

y=1/(3sinx-4cosx+7)

=>y=1/((5(3/5sinx-4/5cosx)+7)

Considering
3/5=cosalpha

So 4/5=sinalpha

And alpha=tan^-1(4/3)

=>y=1/((5(cosalphasinx-sinalphacosx)+7)

=>y=1/((5sin(x-alpha)+7))

We know -1<=sin(x-alpha)<=+1

Hence y will be maximum when sin(x+alpha)=-1

So =y_"max"=1/(5(-1)+7)=1/2

And y will be minimum when sin(x+alpha)=1

So =y_"min"=1/(5*1+7)=1/12