If `f(x)=-1+|x-2|,0<=x<=4` and `g(x)=2-|x|, -1<=x<=3` then find `(fog)(x)` and `(gof)(x)`

f(x)=-1+|x-2|,0<=x<=4,g(x)=2-|x|,-1<=x<=3 Then find fog(x)

If f(x)=x(x-1)(x-2), 0 le x le 4 , then the point xi which satisfies Mean Value Theorem satisfies

If f(x)= {(|1-4x^2|,; 0 lt= x lt 1), ([x^2-2x],; 1 lt= x lt 2):} , where [.] denotes the greatest integer function, then f(x) is

If f(x) = x^2+1/x,xne0 then f(1/x) =

If f(x)=1//x,x=2,deltax=0.2, then df=

Verify Lagranges mean value theorem for f(x)=x(x-1)(x-2) on [0,\ 1/2]

The constant c of Lagrange's theorem for f(x)=x(x-1)(x-2) "in" [0,1//2] is

if f(x)=e^(-1/x^2),x!=0 and f (0)=0 then f'(0) is