if f'(x)=(x -4) (x-5) (x-6) (x-7) then.

Let f(x) =(x-4) (x-5) (x-6) (x-7) then :

Let f(x)=(x-4)(x-5)(x-6)(x-7) then.(A) f'(x)=0 has four roots (B)Three roots of f'(x)=0 lie in (4,5)uu(5,6)uu(6,7) (C) The equation f'(x)=0 has only one real root (D) Three roots of f'(x)=0 lie in (3,4)uu(4,5)uu(5,6)

Let f(x) = ( x-4) (x -5) ( x- 6 ) ( x- 7 ) then

Solve for x: (x-4)/(x-5)+(x-6)/(x-7)=10/3; x!=5,7

If f(x)=3^(x)+4^(x)+5^(x)-6^(x), then f(x)

(x-4)/(x-5)+(x-6)/(x-7)=(10)/(3)

If f (x) = (3x + 4)/( 5x -7), g (x) = (7x +4)/(5x -3) then f [g(x)]=

If f (x) = (3x + 4)/( 5x -7), g (x) = (7x +4)/(5x -3) then f [g(x)]=

If f(x)=(x+1)/((x-2)(x-5)) , then in [4, 6]

Given f(x) is a cubic polynomial in x . If f(x) is divided by (x+3),(x+4),(x+5)and (x+6) , then it leaves the remainders 0, 0 , 4 and 6 respectively . Find the remainder when f(x) is divided by x+7 .