Let f(x) = begin{cases} frac{1}{|x|}, & |x| g | vedclass

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(A) Given $f(x) = \begin{cases} -\frac{1}{x}, & x \le -1 \ ax^2 + b, & -1 < x < 1 \ \frac{1}{x}, & x \ge 1 \end{cases}$.Since $f(x)$ is continuous a

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$-\frac{1}{2}, \frac{3}{2}$

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(A) Given $f(x) = \begin{cases} -\frac{1}{x}, & x \le -1 \ ax^2 + b, & -1 Since $f(x)$ is continuous and differentiable everywhere,it must be continuous and differentiable at $x = 1$ and $x = -1$.For continuity at $x = 1$: $\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) = f(1)$.$a(1)^2 + b = \frac{1}{1} \Rightarrow a + b = 1$.For differentiability at $x = 1$: $f'(x) = \begin{cases} \frac{1}{x^2}, & x  1 \end{cases}$.Left-hand derivative at $x = 1$: $f'(1^-) = 2a(1) = 2a$.Right-hand derivative at $x = 1$: $f'(1^+) = -\frac{1}{(1)^2} = -1$.Since $f'(1^-) = f'(1^+)$,we have $2a = -1 \Rightarrow a = -\frac{1}{2}$.Substituting $a = -\frac{1}{2}$ into $a + b = 1$: $-\frac{1}{2} + b = 1 \Rightarrow b = \frac{3}{2}$.

Let $f(x) = \begin{cases} \frac{1}{|x|}, & |x| \geqslant 1 \\ ax^2 + b, & |x| < 1 \end{cases}$ be continuous and differentiable everywhere. Then $a$ and $b$ are

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