Applied Mathematics

Calculus

Medium

The minimum value of $f(x) = 4x^3 - 48x + 105$ in the interval [1,3] is :

✅ Correct Option: 2

$f'(x) = 12x^2 - 48 = 0 \Rightarrow x^2 = 4 \Rightarrow x = 2$ (in [1, 3]).

Evaluate at critical point and endpoints:

$f(1) = 4 - 48 + 105 = 61$ .

$f(2) = 32 - 96 + 105 = 41$ .

$f(3) = 108 - 144 + 105 = 69$ .

Minimum value = 41 at x = 2.

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The minimum value of f(x)=4x3−48x+105f(x) = 4x^3 - 48x + 105f(x)=4x3−48x+105 in the interval [1,3] is :