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

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The function F(x) is defined as the integral from x² - 2x to 0 of 1/(1+t²) dt. It is established that F(x) is greater than 0 for the interval 0 < x < 2 and less than 0 outside this range. The derivative of F(x) does not vanish outside the interval, indicating that F(x) does not have a minimum. The maximum value of F(x) corresponds to the minimum of the function g(x) = x² - 2x, which occurs at x = 1. Therefore, F(x) has a maximum but no minimum value.
wonguyen1995
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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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