Consider f(x) = begin{cases} tan^{-1}(frac{al | vedclass

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(B) For $f(x)$ to be continuous at $x = \frac{1}{2}$,the left-hand limit,right-hand limit,and the value of the function must be equal: $\lim_{x 	o \f

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$2$

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(B) For $f(x)$ to be continuous at $x = \frac{1}{2}$,the left-hand limit,right-hand limit,and the value of the function must be equal: $\lim_{x 	o \frac{1}{2}^-} 	an^{-1}(\frac{\alpha x + \beta}{\gamma}) = 0$ and $\lim_{x 	o \frac{1}{2}^+} \ln(\beta x^2 + 2) = 0$. From $\ln(\frac{\beta}{4} + 2) = 0$,we get $\frac{\beta}{4} + 2 = 1$,which implies $\beta = -4$. From $	an^{-1}(\frac{\alpha/2 + \beta}{\gamma}) = 0$,we get $\frac{\alpha}{2} + \beta = 0$. Substituting $\beta = -4$,we get $\frac{\alpha}{2} - 4 = 0$,so $\alpha = 8$. For differentiability at $x = \frac{1}{2}$,the derivatives from both sides must be equal: $\frac{d}{dx} [	an^{-1}(\frac{\alpha x + \beta}{\gamma})] = \frac{1}{1 + (\frac{\alpha x + \beta}{\gamma})^2} \cdot \frac{\alpha}{\gamma}$. At $x = \frac{1}{2}$,this is $\frac{\alpha}{\gamma}$ (since $\alpha x + \beta = 0$). $\frac{d}{dx} [\ln(\beta x^2 + 2)] = \frac{2 \beta x}{\beta x^2 + 2}$. At $x = \frac{1}{2}$,this is $\frac{\beta}{-4/4 + 2} = \frac{-4}{1} = -4$. Thus,$\frac{\alpha}{\gamma} = -4 \Rightarrow \frac{8}{\gamma} = -4 \Rightarrow \gamma = -2$. Finally,$\alpha + \beta + \gamma = 8 - 4 - 2 = 2$.

Consider $f(x) = \begin{cases} \tan^{-1}(\frac{\alpha x + \beta}{\gamma}) & x \in (0, \frac{1}{2}) \\ 0 & x = \frac{1}{2} \\ \ln(\beta x^2 + 2) & x \in (\frac{1}{2}, 1) \end{cases}$. If $f(x)$ is continuous and differentiable in its domain,then the value of $\alpha + \beta + \gamma$ is:

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