NCERT Solutions
Class 12 Maths
Application of Derivatives
Ex.6.5 Q.10
Find the maximum value of 2x3 − 24x + 107 in the interval [1, 3].
Find the maximum value of the same function in [−3, −1].
Let f(x) = 2x3 − 24x + 107
So, f’(x) = 6x2 – 24 = 6(x2 – 4)
Now, f’(x) = 0
⟹ 6(x2 – 4) = 0
⟹ x2 – 4 = 0
⟹ x2 = 4
⟹ x = ±2
We first consider the interval [1, 3].
Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the end points of the interval [1, 3].
f(2) = 2 (2)3 – 24 (2) + 107 = 16 − 48 + 107 = 75
f(1) = 2 (1)3 – 24 (1) + 107 = 2 − 24 + 107 = 85
f(3) = 2 (3)3 – 24 (3) + 107 = 54 − 72 + 107 = 89
Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.
Next, we consider the interval [−3, −1].
Evaluate the value of f at the critical point x = −2 ∈ [−3, −1] and at the end points of the
interval [-3, -1].
f(−3) = 2 (−3)3 − 24(−3) + 107 = −54 + 72 + 107 = 125
f(−1) = 2(−1)3 − 24 (−1) + 107 = −2 + 24 + 107 = 129
f(−2) = 2(−2)3 − 24 (−2) + 107 = −16 + 48 + 107 = 139
Hence, the absolute maximum value of f(x) in the interval [−3, −1] is 139 occurring at x = −2