Let ​g(x)equalsleft brace Start 3 By 2 Matrix 1st Row 1st Column x squared plus 3 x 2nd Column ModifyingAbove if x less than 2 With font size decreased by 5 2nd Row 1st Column a 2nd Column if x equals 2 3rd Row 1st Column 4 x plus 7 2nd Column if x greater than 2 EndMatrix
 .
a.
Determine the value of a for which g is continuous from the left at 2.
b.
Determine the value of a for which g is continuous from the right at 2.
c.
Is there a value of a for which g is continuous at 2​?
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Part 1
a. The value of a for which g is continuous from the left at 2 is 
  
0.
​(Simplify your​ answer.)

To answer these questions about the function \( g(x) \), we need to analyze the conditions for continuity from the left, right, and at \( x = 2 \).

The function \( g(x) \) is defined as follows:

\[
g(x) =
\begin{cases}
x^2 + 3x & \text{if } x < 2, \\
a & \text{if } x = 2, \\
4x + 7 & \text{if } x > 2.
\end{cases}
\]

### Part (a): Continuity from the left at \( x = 2 \)

For \( g(x) \) to be continuous from the left at \( x = 2 \), the limit of \( g(x) \) as \( x \) approaches 2 from the left must equal \( g(2) = a \). That is, we need:

\[
\lim_{x \to 2^-} g(x) = a.
\]

For \( x < 2 \), the function is given by \( g(x) = x^2 + 3x \). We will compute the left-hand limit as \( x \to 2 \):

\[
\lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} (x^2 + 3x) = (2^2 + 3(2)) = 4 + 6 = 10.
\]

For \( g(x) \) to be continuous from the left at \( x = 2 \), we need \( a = 10 \).

Thus, the value of \( a \) for which \( g(x) \) is continuous from the left at \( x = 2 \) is \( \boxed{10} \).

### Part (b): Continuity from the right at \( x = 2 \)

For \( g(x) \) to be continuous from the right at \( x = 2 \), the limit of \( g(x) \) as \( x \) approaches 2 from the right must equal \( g(2) = a \). That is, we need:

\[
\lim_{x \to 2^+} g(x) = a.
\]

For \( x > 2 \), the function is given by \( g(x) = 4x + 7 \). We will compute the right-hand limit as \( x \to 2 \):

\[
\lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (4x + 7) = 4(2) + 7 = 8 + 7 = 15.
\]

For \( g(x) \) to be continuous from the right at \( x = 2 \), we need \( a = 15 \).

Thus, the value of \( a \) for which \( g(x) \) is continuous from the right at \( x = 2 \) is \( \boxed{15} \).

### Part (c): Continuity at \( x = 2 \)

For \( g(x) \) to be continuous at \( x = 2 \), both the left-hand and right-hand limits must equal \( g(2) = a \). That is, we need:

\[
\lim_{x \to 2^-} g(x) = \lim_{x \to 2^+} g(x) = g(2) = a.
\]

From Part (a), we found that the left-hand limit is 10, and from Part (b), we found that the right-hand limit is 15. Since these two limits are not equal, there is no single value of \( a \) that makes \( g(x) \) continuous at \( x = 2 \).

Thus, there is **no value** of \( a \) for which \( g(x) \) is continuous at \( x = 2 \).

---

Do you want any details or further clarification?

Here are 5 related questions:
1. What is the definition of continuity at a point?
2. How can you determine if a piecewise function is continuous at a specific point?
3. What happens if the left-hand and right-hand limits of a function are not equal at a point?
4. How do you find limits for piecewise functions?
5. How would the graph of \( g(x) \) behave around \( x = 2 \) for different values of \( a \)?

**Tip:** When solving problems involving continuity, always check both the left-hand and right-hand limits at the point of interest.

Let ​g(x) be defined as a piecewise function: g(x) = {x^2 + 3x, if x < 2; a, if x = 2; 4x + 7, if x > 2}. Determine the value of a for which g is continuous from the left at 2. Determine the value of a for which g is continuous from the right at 2. Is there a value of a for which g is continuous at 2?

Explore how to determine the value of 'a' for which the piecewise function g(x) is continuous at x = 2. This problem involves analyzing left-hand and right-hand limits and applying the continuity theorem. Suitable for high school students in Grades 11-12.

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