Question

Find the maximum and minimum values of 3x⁴ - 8x³+12x² - 48x+1 on the interval [1, 4].

Answer 1

max. value is 257 at x = 4 and min. value is −63 at x = 2

F^|(x) = 12x³ - 24x²+24x - 48 = 0

12(x³ - 2x²+2x - 4) = 0

Since for x = 2, x³ - 2x²+2x - 4 = 0, x - 2 is a factor

On dividing x³ - 2x²+2x - 4 by x - 2, we get,

12(x - 2)(x²+2) = 0

X = 2,4

Now, we shall evaluate the value of f at these points and the end points

F(1) = 3(1)⁴ - 8(1)³+12(1)² - 48(1)+1 = -40

F(2) = 3(2)⁴ - 8(2)³+12(2)² - 48(2)+1 = -63

F(4) = 3(4)⁴ - 8(4)³+12(4)² - 48(4)+1 = 257