If f(x) = tan left(frac{pi}{sqrt{x+1}+4}right | vedclass

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(B) For $f(x)$ to be a real-valued function,the domain requires $x+1 \ge 0$,so $x \ge -1$. As $x$ varies from $-1$ to $\infty$,the term $\sqrt{x+1}$ v

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$(0, 1]$

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(B) For $f(x)$ to be a real-valued function,the domain requires $x+1 \ge 0$,so $x \ge -1$. As $x$ varies from $-1$ to $\infty$,the term $\sqrt{x+1}$ varies from $0$ to $\infty$. Consequently,the denominator $\sqrt{x+1}+4$ varies from $4$ to $\infty$. Thus,the argument of the tangent function,$	heta = \frac{\pi}{\sqrt{x+1}+4}$,varies from $0$ to $\frac{\pi}{4}$. Since the tangent function is strictly increasing in the interval $[0, \frac{\pi}{4}]$,the range of $f(x) = 	an(	heta)$ is $[	an(0), 	an(\frac{\pi}{4})]$. This simplifies to $[0, 1]$. Since none of the options match $[0, 1]$,and assuming the question implies the range of the function as defined,the closest logical interval based on standard function analysis is $(0, 1]$ if $x > -1$.

If $f(x) = \tan \left(\frac{\pi}{\sqrt{x+1}+4}\right)$ is a real-valued function,then the range of $f$ is:

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