Find both the maximum value and the minimum value respectively of 3a4 minus; 8a3 + 12a2 minus; 48a + 25 on the interval [0, 3].

Given, f(a) = 3a⁴ minus; 8a³ + 12a² minus; 48a + 25 fprime;(a) = 12a³ minus; 24a² + 24a minus; 48 = 12(a³ minus;2a² + 2a minus;4) = 12[a²(a minus; 2) + 2(a minus; 2)] = 12(a² + 2)(a minus; 2) For maxima and minima, fprime;(a) = 0 =gt; 12(a² + 2)(a minus; 2)=0 =gt; a = 2, a² = -2 Since a² = -2 is not possible So, a = 2 isin; [0, 3] Now we evaluate the value of f at critical point a = 2 and at the end points of the interval [0, 3] f(0) = 25 f(2) = 3 times; 24 ndash; 8 times; 23 + 12 times; 22 ndash; 48 times; 2 + 25 = 48 ndash; 64 + 48 ndash; 96 + 25 = minus;39 f(3) = 3 times; 34 ndash; 8 times; 33 + 12 times; 32 ndash; 48 times; 3 + 25 = 243 ndash; 216 + 108 minus;144 + 25 = 16 Hence, at a = 0, Maximum value = 25 At a = 2, Minimum value = -39