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(Solved): Let \( A(x)=\int_{0}^{x} f(t) d t \), with \( f(x) \) as in figure. \( \alpha(x) \) has a local min ...






Let \( A(x)=\int_{0}^{x} f(t) d t \), with \( f(x) \) as in figure.
\( \alpha(x) \) has a local minimum on \( (0,6) \) at \(
Let \( A(x)=\int_{0}^{x} f(t) d t \), with \( f(x) \) as in figure. \( \alpha(x) \) has a local minimum on \( (0,6) \) at \( x= \) \( A(x) \) has a local maximum on \( (0,6) \) at \( x= \)


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we have the fundamental theorem, therefore, f(x) = 0 at x=2.5, 4.5 hence, When the derivative crosses the x-axis, that means the function has a loc
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