Question

Find the maximum value of 2x³ – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [–3, –1].

Answer 1

Suppose f(x) = 2x³ – 24 x + 107

∴ f'(x) = 6 x² – 24 = 6(x² – 4)

For the local maxima and local minima we have f'(x) = 0

⇒ 6(x² – 4) = 0

⇒ x² = 4

⇒ x = ±2

As we first consider the interval [1, 3].

Now, 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 (2³)– 24 (2) + 107 = 75

f (1) = 2(1)³ – 24(1) + 107 = 85

f (3) = 2(3)³– 24 (3) + 107 = 89

Thus, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3,

Now, we consider the interval [– 3, – 1].

Then, evaluate the value of f at the critical point x = – 2 ∈ [1, 3]

f (-3) = 2 (-3)³ – 24 (-3) + 107 = 125

f (-2) = 2 (-2)³ – 24 (-3) + 107 = 139

f (-1) = 2 (-1)³ – 24 (-2) + 107 = 129

Thus, the absolute maximum value when x = -2 is 139.