If the function f(x) = frac{1}{2} - tan left( | vedclass

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(B) The domain of $(g \circ f)(x)$ is the set of all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$. First,find the domain of $g(x)

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$\left[ -\frac{1}{2}, \frac{1}{2} \right]$

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(B) The domain of $(g \circ f)(x)$ is the set of all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$. First,find the domain of $g(x) = \sqrt{3 + 4x - 4x^2}$. For $g(x)$ to be defined,$3 + 4x - 4x^2 \ge 0$,which implies $4x^2 - 4x - 3 \le 0$. Factoring the quadratic: $(2x - 3)(2x + 1) \le 0$. This inequality holds for $x \in \left[ -\frac{1}{2}, \frac{3}{2} ight]$. Now,we require $f(x) \in \left[ -\frac{1}{2}, \frac{3}{2} ight]$,so $-\frac{1}{2} \le \frac{1}{2} - 	an \left( \frac{\pi x}{2} ight) \le \frac{3}{2}$. Subtracting $\frac{1}{2}$ from all parts: $-1 \le -	an \left( \frac{\pi x}{2} ight) \le 1$. Multiplying by $-1$ (reversing the inequality): $-1 \le 	an \left( \frac{\pi x}{2} ight) \le 1$. Applying the inverse tangent function: $-\frac{\pi}{4} \le \frac{\pi x}{2} \le \frac{\pi}{4}$. Dividing by $\frac{\pi}{2}$: $-\frac{1}{2} \le x \le \frac{1}{2}$. Since this interval is within the given domain of $f$ $(-1 < x < 1)$,the domain of $(g \circ f)$ is $\left[ -\frac{1}{2}, \frac{1}{2} ight]$.

If the function $f(x) = \frac{1}{2} - \tan \left( \frac{\pi x}{2} \right)$ for $-1 < x < 1$ and $g(x) = \sqrt{3 + 4x - 4x^2}$,then the domain of the composite function $(g \circ f)(x)$ is:

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