Find both the maximum value and the minimum value of 3x⁴ - 8x³ + 12x² - 48x + 25 on the interval [0, 3]

Solution:

Maxima and minima are known as the extrema of a function.

Maxima and minima are the maximum or the minimum value of a function within the given set of ranges

Let f (x) = 3x⁴ - 8x³ + 12x² - 48x + 25

Therefore,

On differentiating wrt x, we get

f' (x) = 12x³ - 24x² + 24x - 48

= 12(x³ - 2x² + 2x - 4)

= 12 [x² (x - 2) + 2(x - 2)]

= 12(x - 2)(x² + 2)

Now,

f' (x) = 0

⇒ x - 2 = 0

⇒ x = 2

Now, we evaluate the value of f at critical point x = 2 and at the end points of the interval [0, 3].

Therefore,

f (2) = 3(2)⁴ - 8(2)³ + 12 (2)² - 48(2) + 25

= 48 - 64 + 48 - 96 + 25

= - 39

f (0) = 3(0)⁴ - 8(0)³ + 12 (0)² - 48(0) + 25

= 25

f (3) = 3(3)⁴ - 8(3)³ + 12 (3)² - 48(3) + 25

= 243 - 216 + 108 - 144 + 25

= 16

Hence, we can conclude that the absolute maximum 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

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 7