The minimum value of f(x) = 3x4 - 8x3 - 48x+25 on [0, 3] isA. 16B. 25C

1.	The minimum value of f(x) = 3x4 - 8x3 - 48x+25 on [0, 3] isA. 16B. 25C. -39D. none of these
Answer» Answer is : C. -39 Given: f(x) = 3x⁴- 8x³- 48x+25. F’(x) = 12x³- 24x²- 48 = 0 F’(x) = 12(x³- 2x²- 4) = 0 Differentiating again, we get, F’’(x) = 3x² – 4x = 0 x(3x – 4) = 0 x = 0 or x = 4/3 Putting the value in equation, we get, f(x) = -39 Hence, C is the correct answer

The minimum value of f(x) = 3x4 - 8x3 - 48x+25 on [0, 3] isA. 16B. 25C. -39D. none of these

Answer»

Answer is : C. -39

Given: f(x) = 3x⁴- 8x³- 48x+25.

F’(x) = 12x³- 24x²- 48 = 0

F’(x) = 12(x³- 2x²- 4) = 0

Differentiating again, we get,

F’’(x) = 3x² – 4x = 0

x(3x – 4) = 0

x = 0 or x = 4/3

Putting the value in equation, we get,

f(x) = -39

Hence, C is the correct answer