Math Problem Statement
Reponse
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.
Solution
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 at . The function is defined as:
\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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Limits
Piecewise Functions
Rational Functions
Formulas
Limit of f(x) as x approaches a
Continuity condition: lim x->a f(x) = f(a)
Quadratic function: f(x) = ax^2 + bx + c
Theorems
Continuity Theorem
Limit Theorem
Suitable Grade Level
Grades 11-12 (Pre-Calculus/Calculus)
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