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The minimum value of the function $f(x)=x^3-12 x$ on the interval [0, 3] is: |
0 -9 -16 -19 |
-16 |
The correct answer is Option (3) → -16 $f(x)=x^3-12x$ $f'(x)=3x^2-12=3(x^2-4)$ $f'(x)=0 \Rightarrow x=\pm 2$ $x=2 \in [0,3]$ $f(0)=0,\;\; f(2)=8-24=-16,\;\; f(3)=27-36=-9$ $\text{Minimum value} = -16$ The minimum value is $-16$. |
