If f(x) = begin{cases} frac{1}{|x|} & ; |x| g | vedclass

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(D) Given $f(x) = \begin{cases} \frac{1}{|x|} & ; |x| \geq 1 \ ax^2 + b & ; |x| < 1 \end{cases}$.For $x \geq 1$,$f(x) = \frac{1}{x}$. For $x \leq -1$

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$-1/2, 3/2$

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(D) Given $f(x) = \begin{cases} \frac{1}{|x|} & ; |x| \geq 1 \ ax^2 + b & ; |x| For $x \geq 1$,$f(x) = \frac{1}{x}$. For $x \leq -1$,$f(x) = -\frac{1}{x}$.Since $f(x)$ is differentiable at every point,it must be continuous at $x = 1$.$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) \implies a(1)^2 + b = \frac{1}{1} \implies a + b = 1 \quad \dots(1)$.Also,$f(x)$ must be differentiable at $x = 1$.$f'(x) = \begin{cases} -\frac{1}{x^2} & ; x > 1 \ 2ax & ; -1 Equating the derivatives at $x = 1$:$\lim_{x 	o 1^-} f'(x) = \lim_{x 	o 1^+} f'(x) \implies 2a(1) = -\frac{1}{(1)^2} \implies 2a = -1 \implies a = -\frac{1}{2}$.Substituting $a = -\frac{1}{2}$ into equation $(1)$:$-\frac{1}{2} + b = 1 \implies b = 1 + \frac{1}{2} = \frac{3}{2}$.Thus,the values are $a = -\frac{1}{2}$ and $b = \frac{3}{2}$.

If $f(x) = \begin{cases} \frac{1}{|x|} & ; |x| \geq 1 \\ ax^2 + b & ; |x| < 1 \end{cases}$ is differentiable at every point of the domain,then the values of $a$ and $b$ are respectively

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