If a function f:(-1,1) rightarrow B(subseteq | vedclass

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(B) Given, $f(x) = x + x^2 + x^3 + \ldots \infty$. This is an infinite geometric series with first term $a = x$ and common ratio $r = x$. Since the do

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

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(B) Given, $f(x) = x + x^2 + x^3 + \ldots \infty$. This is an infinite geometric series with first term $a = x$ and common ratio $r = x$. Since the domain is $(-1, 1)$, $|x| To find the range $B$, we set $y = \frac{x}{1-x}$. $y(1-x) = x \Rightarrow y - xy = x \Rightarrow y = x(1+y) \Rightarrow x = \frac{y}{1+y}$. Since $-1 Case $1$: $\frac{y}{1+y} > -1 \Rightarrow \frac{y + 1 + y}{1+y} > 0 \Rightarrow \frac{2y+1}{1+y} > 0$. Critical points are $y = -1$ and $y = -1/2$. Testing intervals, we get $y \in (-\infty, -1) \cup (-1/2, \infty)$. Case $2$: $\frac{y}{1+y}  0 \Rightarrow y > -1$. Taking the intersection of Case $1$ and Case $2$, we get $y \in (-1/2, \infty)$. Thus, $B = (-1/2, \infty)$.

If a function $f:(-1,1) \rightarrow B(\subseteq R)$ is defined as $f(x)=x+x^2+x^3+\ldots \infty$, then in order to have the inverse function of $f$, $B$ is equal to

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