Let f(x) = log(1 + x^2) and A be a constant s | vedclass

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(A) By the Mean Value Theorem,for any $x 
eq y$,there exists a $c$ between $x$ and $y$ such that $\frac{f(x) - f(y)}{x - y} = f'(c)$.Given $f(x) = \l

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equal to $1$

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(A) By the Mean Value Theorem,for any $x 
eq y$,there exists a $c$ between $x$ and $y$ such that $\frac{f(x) - f(y)}{x - y} = f'(c)$.Given $f(x) = \log(1 + x^2)$,we find the derivative $f'(x) = \frac{2x}{1 + x^2}$.To find the range of $f'(x)$,we analyze $g(x) = \frac{2x}{1 + x^2}$.Setting $g'(x) = \frac{2(1 + x^2) - 2x(2x)}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} = 0$,we get $x = \pm 1$.The maximum value of $f'(x)$ is $f'(1) = \frac{2(1)}{1 + 1^2} = 1$ and the minimum value is $f'(-1) = \frac{2(-1)}{1 + (-1)^2} = -1$.Thus,$|f'(c)| \leq 1$ for all $c \in \mathbb{R}$.Since $\frac{|f(x) - f(y)|}{|x - y|} = |f'(c)| \leq 1$,the least possible value of $A$ is $1$.

Let $f(x) = \log(1 + x^2)$ and $A$ be a constant such that $\frac{|f(x) - f(y)|}{|x - y|} \leq A$ for all real $x, y$ where $x \neq y$. Then,the least possible value of $A$ is

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