If f(x) =2x-1 and g(x)=3x+2 then find (fog) (x) a)2(3x+1) b)2(3x+2) c)(2x+1) d)3(3x+1)

Options

A. 2(3x+1)
B. 2(3x+2)
C. 3(2x+1)
D. 3(3x+1)

Answer

[-1,1]

Explanation

`g(x+y)=g(x)+g(y)+3x^(2)y+3xy^(2)" (1)"`
`"or "g'(x+y)=g'(x)+6yx+3y^(2)" (differentiating w.r.t. x keeping y as constant)"`
Put `x=0.` Then,
`"or "g'(y)=g'(0)+3y^(2)=-4+3y^(2)`
`"or "g'(x)=-4+3x^(2)`
`"or "g(x)=-4x+x^(3)+c`
Now, put `x=y=0 in (1)." Then "g(0)=g(0)+g(0)+0`
`"or "g(0)=0`
`"or "g(x)=x^(3)-4x`
`g(x)=0rArrx^(3)-4x=0rArrx=0,2,-2.`
Hence. three roots.
`sqrt(g(x))=sqrt(x^(3)-4x)" is defined if "x^(3)-4xge0`
`"or "x in [-2, 0]cup[2,oo).`
`"Also, "g'(x)=3x^(2)-4rArrg'(1)=-1.`