Find both the maximum value and the minimum value of
3x4 – 8x3 + 12x2 – 48x + 25
Let f (x) = 3x4 – 8x3 + 12x2 – 48x + 25
f ' (x) = 12x3 – 24x2 + 24x – 48
f ' (x) = 0 ⇒ 12x3 – 24x2 + 24x – 48 = 0 ⇒ x3 – 2x2 + 2x – 4 = 0
⇒ (x – 2) (x2 + 2) = 0 ⇒ x = 2 , ± i .
Now x = 2 is the only real value in [0, 3]
f (0) = 0 – 0 + 0 – 0 + 25 = 25
f (2) = 48 – 64 + 48 – 96 + 25 = – 39
f (3) = 243 – 216 + 108 – 144 + 25 = 16
∴ max. value = 25 and min. value = – 39.
Find the shortest distance of the point (0, c) from the parabola y = x2 , where 0 ≤ c ≤ 5.