Let f(x)=3x^4-8x^3+12x^2-48x+25 f^’ (x)=12x^3-24x^2+24x-48 =12(x^3-2x^2+2x-4) =12[x^2 (x-2)+2(x-2)] =12(x-2)(x^2+2) Now, f^’ (x)=0 gives x =2 or x^2+2=0 for which there are real roots Therefore, we consider only x=2∈[0,3] Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3]. f(2)=3〖(2)〗^4-8〖(2)〗^3+12〖(2)〗^2-48(2)+25 =3(16)-8(8)+12(4)-48(2)+25=48-64+48-96+25 =-39 f(0)=3〖(0)〗^4-8〖(0)〗^3+12〖(0)〗^2-48(0)+25=3(0)-8(0)+12(0)-48(0)+25=25 f(3)=3〖(3)〗^4-8〖(3)〗^3+12〖(3)〗^2-48(3)+25 F(3)=3(81)-8(27)+12(9)-48(3)+25 =243-216+108-144+25 =16 Hence, we can conclude that the absolute value of f on [0, 3] is 25 occurring at x=0 and the absolute minimum value of f at [0, 3] is – 39 occurring at x = 2.