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Let's solve the given problem step by step.

We are given the piecewise function:

\[
f(x) = 
\begin{cases}
\frac{3x^2 - 4x - 4}{x - 2}, & x < 2 \\
2\sqrt{x}, & x \geq 2
\end{cases}
\]

### Part (a) - Calculating the one-sided limits at \( x = 2 \) and determining continuity.

#### 1. **Right-hand limit** as \( x \to 2^+ \):

For \( x \geq 2 \), the function is defined as \( f(x) = 2\sqrt{x} \). To find the limit from the right at \( x = 2 \):

\[
\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} 2\sqrt{x} = 2\sqrt{2} = 2\sqrt{2}
\]

#### 2. **Left-hand limit** as \( x \to 2^- \):

For \( x < 2 \), the function is \( f(x) = \frac{3x^2 - 4x - 4}{x - 2} \).

We need to compute:

\[
\lim_{x \to 2^-} \frac{3x^2 - 4x - 4}{x - 2}
\]

The numerator is \( 3x^2 - 4x - 4 \). Let's factorize this polynomial. Using the quadratic formula:

\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-4)}}{2(3)} = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6}
\]

Thus, the roots are \( x = 2 \) and \( x = -\frac{2}{3} \). So the factorization of the quadratic becomes:

\[
3x^2 - 4x - 4 = 3(x - 2)\left(x + \frac{2}{3}\right)
\]

Now, we can simplify the limit expression:

\[
\lim_{x \to 2^-} \frac{3(x - 2)\left(x + \frac{2}{3}\right)}{x - 2}
\]

Canceling out the \( (x - 2) \) terms:

\[
\lim_{x \to 2^-} 3\left(x + \frac{2}{3}\right) = 3\left(2 + \frac{2}{3}\right) = 3 \times \frac{8}{3} = 8
\]

#### 3. **Conclusion about continuity** at \( x = 2 \):

The right-hand limit is \( 2\sqrt{2} \) and the left-hand limit is \( 8 \). Since:

\[
\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)
\]

The function is **not continuous** at \( x = 2 \).

#### 4. **Is the function piecewise continuous?**

Even though the function is not continuous at \( x = 2 \), it is continuous on the intervals \( (-\infty, 2) \) and \( [2, \infty) \), meaning it is piecewise continuous on \( \mathbb{R} \).

---

Would you like any details or further clarifications?

### Related Questions:
1. What are the conditions for a function to be continuous at a point?
2. How does factoring polynomials help in simplifying limits?
3. Can a function be discontinuous at a point but still be integrable?
4. What are some examples of piecewise functions that are continuous at their boundary points?
5. How can we graphically represent the one-sided limits and continuity of this function?

### Tip:
When finding the limit of a rational function, always check for common factors in the numerator and denominator to simplify the expression before substituting the limit value.

Calculate the one-sided limits at x = 2 and determine whether the function is continuous at x = 2. Is the function piecewise continuous in R?

In this problem, we calculate the one-sided limits at x = 2 for a piecewise function and determine if the function is continuous at that point. The solution involves algebraic simplification and limit calculation.

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