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Let's work through the exercises one by one based on the image you have uploaded.

### Exercice 1:
We are asked to study the continuity of the function \( f(x) \) at \( a = 1 \). The function is defined as:

\[
f(x) =
\begin{cases}
\frac{3x^2 - x - 2}{14x - 3} & \text{if } x \neq 1 \\
\frac{5}{4} & \text{if } x = 1
\end{cases}
\]

To check if the function is continuous at \( a = 1 \), we need to verify the following:

1. **Existence of \( f(1) \)**: Given in the problem, \( f(1) = \frac{5}{4} \).
   
2. **Limit of \( f(x) \) as \( x \to 1 \)**:
   \[
   \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{3x^2 - x - 2}{14x - 3}
   \]
   We need to simplify this expression at \( x = 1 \).

3. **Check if the limit matches the value of \( f(1) \)**:
   \[
   \lim_{x \to 1} f(x) = f(1) = \frac{5}{4}
   \]
   
### Exercice 2:
We are given the function \( g(x) \), defined as:

\[
g(x) =
\begin{cases}
\frac{x^2 + x - 6}{\sqrt{x^2 + 12x - 4}} & \text{if } x > 2 \\
\frac{x^2 + ax + a + 1}{x^2 - 2} & \text{if } x < 2 \\
l & \text{if } x = 2
\end{cases}
\]

The question asks us to determine if there exist values of \( a \) and \( l \) such that the function \( g(x) \) is continuous at \( x = 2 \).

To solve this:

1. **Existence of \( g(2) = l \)**: We need to determine \( l \), which will be the value of the function at \( x = 2 \).

2. **Limit of \( g(x) \) as \( x \to 2^+ \)** (from the right):
   \[
   \lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} \frac{x^2 + x - 6}{\sqrt{x^2 + 12x - 4}}
   \]

3. **Limit of \( g(x) \) as \( x \to 2^- \)** (from the left):
   \[
   \lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} \frac{x^2 + ax + a + 1}{x^2 - 2}
   \]

4. **Check if the left and right limits match**: We find \( a \) and \( l \) such that the limits from the left and right, and \( g(2) = l \), are equal.

Let me compute the required limits for both exercises and determine the values of \( a \) and \( l \) in Exercise 2.### Exercice 1:
We calculated the limit of \( f(x) \) as \( x \to 1 \):
\[
\lim_{x \to 1} f(x) = 0
\]
Since \( f(1) = \frac{5}{4} \), and the limit as \( x \to 1 \) does not match the value of the function at \( x = 1 \), the function \( f(x) \) is **not continuous** at \( x = 1 \).

### Exercice 2:
We need to find values of \( a \) and \( l \) that make the function \( g(x) \) continuous at \( x = 2 \).

1. The limit of \( g(x) \) as \( x \to 2^+ \) (from the right) is:
   \[
   \lim_{x \to 2^+} g(x) = 0
   \]

2. The limit of \( g(x) \) as \( x \to 2^- \) (from the left) is:
   \[
   \lim_{x \to 2^-} g(x) = \frac{3a}{2} + \frac{5}{2}
   \]

3. To ensure continuity, we set the two limits equal and solve for \( a \) and \( l \):
   \[
   \frac{3a}{2} + \frac{5}{2} = 0 \quad \Rightarrow \quad a = -\frac{5}{3}
   \]

   We also find that \( l = 0 \).

Thus, the function \( g(x) \) is continuous at \( x = 2 \) if \( a = -\frac{5}{3} \) and \( l = 0 \).

---

Would you like further details or have any questions?

Here are 5 related questions to expand on these topics:
1. What is the general condition for a function to be continuous at a point?
2. How do we handle limits involving square roots when calculating continuity?
3. Can a function be discontinuous but still differentiable at a point?
4. What other types of discontinuities exist besides the one found in \( f(x) \)?
5. How can we extend this analysis to piecewise-defined functions?

**Tip**: When checking the continuity of a piecewise function, always calculate the left-hand and right-hand limits separately, and compare them with the function's value at that point.

Étudier la continuité de la fonction f(x) et déterminer les valeurs de a et l pour que g(x) soit continue en x=2.

Explore how to study the continuity of a piecewise function f(x) and determine the values of parameters a and l for the function g(x) to be continuous at x=2. Learn the necessary steps involving limits, rational functions, and continuity conditions.

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