Given f(x) = frac{1}{2} - tan^{-1}left(frac{p | vedclass

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(A) The domain of $(f + g)$ is the intersection of the domains of $f(x)$ and $g(x)$.For $f(x) = \frac{1}{2} - 	an^{-1}\left(\frac{\pi x}{2}ight)$,t

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$ \left(-\frac{1}{2}, 1\right) $

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(A) The domain of $(f + g)$ is the intersection of the domains of $f(x)$ and $g(x)$.For $f(x) = \frac{1}{2} - 	an^{-1}\left(\frac{\pi x}{2}ight)$,the given domain is $-1 For $g(x) = \sqrt{3 + 4x - 4x^2}$ to be defined,we must have $3 + 4x - 4x^2 \geq 0$.Multiplying by $-1$,we get $4x^2 - 4x - 3 \leq 0$.Factoring the quadratic: $(2x + 1)(2x - 3) \leq 0$.This inequality holds for $x \in \left[-\frac{1}{2}, \frac{3}{2}ight]$.The domain of $(f + g)$ is the intersection of $(-1, 1)$ and $\left[-\frac{1}{2}, \frac{3}{2}ight]$.Intersection: $\left(-\frac{1}{2}, 1ight)$.

$x$	$-4$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$	$4$
$y = f(x) = x^2$

Given $f(x) = \frac{1}{2} - \tan^{-1}\left(\frac{\pi x}{2}\right)$ for $-1 < x < 1$ and $g(x) = \sqrt{3 + 4x - 4x^2}$. Find the domain of $(f + g)$.

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