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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Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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