Let f(x) = min {x, x^2} for every real number | vedclass

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(A) The function is defined as $f(x) = \min \{x, x^2\}$.By comparing $x$ and $x^2$,we find that $x^2 \leq x$ when $0 \leq x \leq 1$,and $x < x^2$ othe

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$f(x)$ is continuous for all $x$

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(A) The function is defined as $f(x) = \min \{x, x^2\}$.By comparing $x$ and $x^2$,we find that $x^2 \leq x$ when $0 \leq x \leq 1$,and $x Thus,$f(x) = \begin{cases} x, & x  1 \end{cases}$.Checking continuity at $x = 0$: $\lim_{x 	o 0^-} f(x) = 0$ and $\lim_{x 	o 0^+} f(x) = 0^2 = 0$. Since $f(0) = 0$,it is continuous at $x = 0$.Checking continuity at $x = 1$: $\lim_{x 	o 1^-} f(x) = 1^2 = 1$ and $\lim_{x 	o 1^+} f(x) = 1$. Since $f(1) = 1$,it is continuous at $x = 1$.Now,checking differentiability: $f'(x) = \begin{cases} 1, & x  1 \end{cases}$.At $x = 0$: Left derivative is $1$,right derivative is $2(0) = 0$. Since $1 
eq 0$,it is not differentiable at $x = 0$.At $x = 1$: Left derivative is $2(1) = 2$,right derivative is $1$. Since $2 
eq 1$,it is not differentiable at $x = 1$.Thus,$f(x)$ is continuous for all $x$.

Let $f(x) = \min \{x, x^2\}$ for every real number $x$. Then:

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