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GUJCET · Maths · Application of Derivatives
The absolute minimum value of the function \(f(x)=x^3-18 x^2+96 x, x \in[0,9]\) is ___________ .
- A -160
- B 0
- C -135
- D 126
Answer & Solution
Correct Answer
(B) 0
Step-by-step Solution
Detailed explanation
\(f'(x) = 3x^2 - 36x + 96\) \(3x^2 - 36x + 96 = 0 \Rightarrow 3(x-4)(x-8) = 0 \Rightarrow x=4, 8\)
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