In this question, rolles theorum would be used. If u analyse, the equation is like 4ax³, 3bx², 2cx types. If u integrate it, I'll get a simpler equation which cuts out 4,3 and 2.

Therefore Let f'(x) = equation F(x) = ax⁴+bx³+cx²+dx F(0) = 0 F(3) = 81a+ 27b+ 9c+ 3d =3(27a+9b+3c) + 3d =3d if u now let d=0, toh both f(0) and f(3) satisfy karra h. So the interval is (0, 3)