Minimum Value of F(x): Does It Have a Maximum?

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SUMMARY

The function F(x) defined as F(x) = ∫(x² - 2x)⁰ (1/(1+t²)) dt is analyzed for its minimum and maximum values. It is established that F(x) is positive for the interval 0 < x < 2 and negative outside this range. The derivative of F(x) does not vanish outside the interval, indicating that there is no minimum value. The maximum occurs at the minimum of the function g(x) = x² - 2x, which is achieved at x = 1.

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Find the minimum value of F(x)=\int_{x^2-2x}^{0} 1/(1+t^2)\,d
Does it has a maximum value? why?
 
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wonguyen1995 said:
Find the minimum value of F(x)=\int_{x^2-2x}^{0} 1/(1+t^2)\,d
Does it has a maximum value? why?

Defining the function as...

$\displaystyle f(x) = \int_{x^{2} - 2 x}^{0} \frac{d t } {1 + t^{2}}\ (1)$

... it is easy to see that is f(x)> 0 for 0 < x < 2 and f(x) < 0 outside this interval. Since the derivative of f(x) outside the interval 0 < x < 2 never vanishes, f(x) don't have minimum. The maximum of the function is the minimum of the function $\displaystyle g(x) = x^{2} - 2\ x$ and it happens for x=1...

Kind regards

$\chi$ $\sigma$
 
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