$$ x = \left\{ \begin{array} { l } { x ^ { 2 } - x ^ { 3 } , x - 1 } \end{array} ight. $$

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$$ x = \left\{ \begin{array} { l } { x ^ { 2 } - x ^ { 3 } , x < - 1 } \ { 1 , x = - 1 } \ { \frac { x ^ { 2 } + 2 x - 1 } { x } , x > - 1 } \end{array} ight. $$

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We are given the function $$ f(x)= \begin{cases} x^{2}-x^{3}, & x<-1,\[1mm] 1, & x=-1,\[1mm] \frac{x^{2}+2x-1}{x}, & x>-1. \end{cases} $$ Below we analyze its properties step by step. --- **1. Analysis at $$ x=-1 $$:** *Step 1: Left-hand limit as $$ x \to -1^- $$* For $$ x<-1 $$ the function is $$ f(x)=x^2-x^3 $$. Compute the limit by direct substitution (since this is a polynomial): $$ \lim_{x\to -1^-} f(x)=(-1)^2-(-1)^3=1-(-1)=2. $$ *Step 2: Right-hand limit as $$ x \to -1^+ $$* For $$ x>-1 $$ the function is $$ f(x)=\frac{x^{2}+2x-1}{x}. $$ Substitute $$ x=-1 $$ into the expression (even though the piece is defined only for $$ x>-1 $$, we use this to compute the limit): $$ \lim_{x\to -1^+} f(x)=\frac{(-1)^2+2(-1)-1}{-1}=\frac{1-2-1}{-1}=\frac{-2}{-1}=2. $$ *Step 3: Compare with the defined value at $$ x=-1 $$* We have $$ f(-1)=1. $$ Since both one‐sided limits equal $$ 2 $$ but $$ f(-1)=1 $$, the function has a discontinuity at $$ x=-1 $$. In fact, the limit exists but does not equal the defined value. Such a discontinuity is removable if we choose to redefine $$ f(-1) $$ as $$ 2 $$. --- **2. Analysis for $$ x > -1 $$:** The expression is $$ f(x)=\frac{x^{2}+2x-1}{x}. $$ *Step 1: Simplify the expression (for $$ x\neq 0 $$)* Divide each term by $$ x $$: $$ \frac{x^{2}+2x-1}{x}=x+2-\frac{1}{x}. $$ *Step 2: Identify the domain* Since the third piece applies for $$ x > -1 $$, it includes all values greater than $$-1$$ except that the expression is not defined at $$ x=0 $$ (because of the denominator $$ x $$). Hence, even though $$ x=0 $$ satisfies $$ x>-1 $$, the simplified expression shows that there is a vertical asymptote at $$ x=0 $$. --- **3. Analysis for $$ x < -1 $$:** For $$ x<-1 $$ the function $$ f(x)=x^{2}-x^{3} $$ is a polynomial and is continuous for all $$ x $$ in its domain (here, $$ x<-1 $$). --- **Summary of Findings:** - **At $$ x=-1 $$:** The left-hand and right-hand limits are both $$ 2 $$ but $$ f(-1)=1 $$. The function is discontinuous at $$ x=-1 $$; the discontinuity is removable by redefining $$ f(-1)=2 $$. - **For $$ x > -1 $$:** The function is given by $$ f(x)=x+2-\frac{1}{x} $$ but is not defined at $$ x=0 $$. Therefore, there is a vertical asymptote at $$ x=0 $$. - **For $$ x<-1 $$:** The function $$ f(x)=x^{2}-x^{3} $$ is continuous. Thus, the overall function is continuous on $$ (-\infty,-1) $$ and on $$ (-1,0) \cup (0, \infty) $$ (with a vertical asymptote at $$ x=0 $$), but it is discontinuous at $$ x=-1 $$ since $$ f(-1) $$ does not equal the limit $$ 2 $$. The function has a discontinuity at $$ x = -1 $$ because the left-hand and right-hand limits are both 2, but $$ f(-1) = 1 $$. For $$ x > -1 $$, there's a vertical asymptote at $$ x = 0 $$. The function is continuous for all other values of $$ x $$.

\( x = \left\{ \begin{array} { l } { x ^ { 2 } - x ^ { 3 } , x < - 1 } \\ { 1 , x = - 1 } \\ { \frac { x ^ { 2 } + 2 x - 1 } { x } , x > - 1 } \end{array} \right. \)

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