Question Statement

Find the points of discontinuity for the piecewise function:

$f(x)=\{\begin{array}{ll}x& ,x<0\\ x^2& ,0\le x\le 2\\ x& ,x\ge 2\end{array}$

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Background and Explanation

A function $f(x)$ is continuous at a point $x=a$ if the left-hand limit, right-hand limit, and function value all exist and are equal: ${\lim }_{x\to a^{-}}f(x)={\lim }_{x\to a^{+}}f(x)=f(a)$. For piecewise functions, we must check continuity at the boundary points where the function definition changes.

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Solution

To determine where $f(x)$ is discontinuous, we examine the boundary points $x=0$ and $x=2$ where the function definition changes.

Checking Continuity at $x=0$

Left-hand limit: As $x$ approaches $0$ from the left ($x<0$), we use $f(x)=x$: ${\lim }_{x\to 0^{-}}f(x)={\lim }_{x\to 0^{-}}x=0$

Right-hand limit: As $x$ approaches $0$ from the right ($0\le x\le 2$), we use $f(x)=x^2$: ${\lim }_{x\to 0^{+}}f(x)={\lim }_{x\to 0^{+}}{x}^2=0$

Function value: At $x=0$, using the middle piece $f(x)=x^2$: $f(0)=0^2=0$

Conclusion: Since ${\lim }_{x\to 0^{-}}f(x)={\lim }_{x\to 0^{+}}f(x)=f(0)=0$, the function is continuous at $x=0$.

Checking Continuity at $x=2$

Left-hand limit: As $x$ approaches $2$ from the left ($0\le x\le 2$), we use $f(x)=x^2$: ${\lim }_{x\to 2^{-}}f(x)={\lim }_{x\to 2^{-}}{x}^2=2^2=4$

Right-hand limit: As $x$ approaches $2$ from the right ($x\ge 2$), we use $f(x)=x$: ${\lim }_{x\to 2^{+}}f(x)={\lim }_{x\to 2^{+}}x=2$

Function value: At $x=2$, using the middle piece $f(x)=x^2$: $f(2)=2^2=4$

Conclusion: Since ${\lim }_{x\to 2^{-}}f(x)=4$ and ${\lim }_{x\to 2^{+}}f(x)=2$, we have: ${\lim }_{x\to 2^{-}}f(x)\ne {\lim }_{x\to 2^{+}}f(x)$

Therefore, the function is discontinuous at $x=2$.

Final Answer

The point of discontinuity is $x=2$.

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Key Formulas or Methods Used

• Continuity Condition: $f(x)$ is continuous at $x=a$ if $\underset{x\to a^{-}}{\lim }f(x)=\underset{x\to a^{+}}{\lim }f(x)=f(a)$
• Left-hand Limit: $\underset{x\to a^{-}}{\lim }f(x)$ — limit as $x$ approaches $a$ from values less than $a$
• Right-hand Limit: $\underset{x\to a^{+}}{\lim }f(x)$ — limit as $x$ approaches $a$ from values greater than $a$
• Piecewise Function Analysis: Evaluating limits using the appropriate function definition for each interval

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Summary of Steps

1. Identify boundary points where the piecewise definition changes (at $x=0$ and $x=2$)
2. Calculate left-hand limit at each boundary point using the function piece defined for $x<a$
3. Calculate right-hand limit at each boundary point using the function piece defined for $x>a$
4. Evaluate the function at each boundary point
5. Compare the three values: if left limit = right limit = function value, the function is continuous; otherwise, it is discontinuous
6. Conclude which points are points of discontinuity (only $x=2$ in this case)