| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if the function \( f(x) = \frac{4x+5}{9-3x} \) is continuous at \( x = -1 \), we need to check three conditions for continuity at that point: 1. \( f(-1) \) exists. 2. \( \lim_{x \to -1} f(x) \) exists. 3. \( \lim_{x \to -1} f(x) = f(-1) \). Let's check these conditions: 1. **Existence of \( f(-1) \):** Substitute \( x = -1 \) into \( f(x) \): \[ f(-1) = \frac{4(-1) + 5}{9 - 3(-1)} = \frac{-4 + 5}{9 + 3} = \frac{1}{12} \] So, \( f(-1) \) exists and is equal to \( \frac{1}{12} \). 2. **Existence of \( \lim_{x \to -1} f(x) \):** To find this limit, we can directly substitute \( x = -1 \) into the function, as long as the denominator is not zero at \( x = -1 \). Since we already evaluated \( f(-1) \), and the denominator was not zero, we can say: \[ \lim_{x \to -1} f(x) = \frac{1}{12} \] So, the limit exists. 3. **Equality of \( \lim_{x \to -1} f(x) \) and \( f(-1) \):** Since both \( \lim_{x \to -1} f(x) \) and \( f(-1) \) are equal to \( \frac{1}{12} \), they are equal. Since all three conditions are satisfied, we can conclude that the function \( f(x) = \frac{4x+5}{9-3x} \) is continuous at \( x = -1 \). |
True |
| deepseekr1 | The function \( f(x) = \frac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \) if three conditions are met: 1. \( f(-1) \) is defined. 2. \( \lim_{x \to -1} f(x) \) exists. 3. \( \lim_{x \to -1} f(x) = f(-1) \). **Step 1: Check if \( f(-1) \) is defined.** Substitute \( x = -1 \) into the function: \[ f(-1) = \frac{4(-1) + 5}{9 - 3(-1)} = \frac{-4 + 5}{9 + 3} = \frac{1}{12}. \] The denominator \( 9 - 3(-1) = 12 \neq 0 \), so \( f(-1) \) is defined. **Step 2: Compute \( \lim_{x \to -1} f(x) \).** Since the denominator at \( x = -1 \) is non-zero, the function is continuous at this point (polynomials are continuous everywhere, and their quotients are continuous where the denominator is non-zero). Thus: \[ \lim_{x \to -1} f(x) = f(-1) = \frac{1}{12}. \] **Step 3: Compare the limit to \( f(-1) \).** Both \( \lim_{x \to -1} f(x) \) and \( f(-1) \) equal \( \frac{1}{12} \). **Conclusion:** All three conditions are satisfied. Therefore, \( f(x) \) is **continuous at \( x = -1 \)**. **Final Answer:** \boxed{Yes} |
True |
| gpto1 | To determine if the function \( f(x) = \dfrac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \), we need to check the following three conditions for continuity at that point: 1. **The function is defined at \( x = -1 \).** 2. **The limit of the function exists as \( x \) approaches \( -1 \).** 3. **The limit of the function as \( x \) approaches \( -1 \) equals the function's value at \( x = -1 \).** Let's examine each of these steps in detail. --- ### **1. The function is defined at \( x = -1 \).** First, evaluate \( f(-1) \): \[ \begin{align*} f(-1) &= \dfrac{4(-1) + 5}{9 - 3(-1)} \\ &= \dfrac{-4 + 5}{9 + 3} \\ &= \dfrac{1}{12} \end{align*} \] Since \( f(-1) = \dfrac{1}{12} \) is a real number, the function is defined at \( x = -1 \). --- ### **2. The limit of the function exists as \( x \) approaches \( -1 \).** Compute the limit of \( f(x) \) as \( x \) approaches \( -1 \): \[ \lim_{x \to -1} f(x) = \lim_{x \to -1} \dfrac{4x + 5}{9 - 3x} \] Since the denominator \( 9 - 3x \) does not equal zero at \( x = -1 \) (as calculated earlier, it equals \( 12 \)), and both the numerator and the denominator are continuous functions (polynomials are continuous everywhere), the function \( f(x) \) is continuous wherever it is defined, except possibly where the denominator is zero. Therefore, the limit exists and is equal to the function evaluated at \( x = -1 \): \[ \lim_{x \to -1} f(x) = f(-1) = \dfrac{1}{12} \] --- ### **3. The limit equals the function's value at \( x = -1 \).** From the previous steps: \[ \lim_{x \to -1} f(x) = \dfrac{1}{12} \quad \text{and} \quad f(-1) = \dfrac{1}{12} \] Since the limit of \( f(x) \) as \( x \) approaches \( -1 \) equals the function's value at that point, the third condition for continuity is satisfied. --- ### **Conclusion** All three conditions are met: 1. \( f(-1) \) is defined. 2. \( \lim_{x \to -1} f(x) \) exists. 3. \( \lim_{x \to -1} f(x) = f(-1) \). **Therefore, the function \( f(x) = \dfrac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \).** |
True |
| GPT-4o | To determine if the function \( f(x) = \frac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \), we need to examine three main criteria for continuity at a point \( x = a \): 1. The function \( f(x) \) is defined at \( x = a \). 2. The limit of \( f(x) \) as \( x \) approaches \( a \) exists. 3. The limit of \( f(x) \) as \( x \) approaches \( a \) is equal to \( f(a) \). Let's go through these steps for \( x = -1 \): ### Step 1: Is \( f(x) \) defined at \( x = -1 \)? First, we need to check if \( f(-1) \) is defined. \[ f(-1) = \frac{4(-1) + 5}{9 - 3(-1)} = \frac{-4 + 5}{9 + 3} = \frac{1}{12} \] Since \( f(-1) \) exists and equals \( \frac{1}{12} \), the function is defined at \( x = -1 \). ### Step 2: Does the limit of \( f(x) \) as \( x \) approaches \( -1 \) exist? We need to find: \[ \lim_{x \to -1} f(x) \] \[ \lim_{x \to -1} \frac{4x + 5}{9 - 3x} \] By direct substitution: \[ \lim_{x \to -1} \frac{4(-1) + 5}{9 - 3(-1)} = \frac{-4 + 5}{9 + 3} = \frac{1}{12} \] The limit exists and is equal to \( \frac{1}{12} \). ### Step 3: Is the limit of \( f(x) \) as \( x \) approaches \( -1 \) equal to \( f(-1) \)? We have already computed both: \[ f(-1) = \frac{1}{12} \] and \[ \lim_{x \to -1} f(x) = \frac{1}{12} \] Since \[ \lim_{x \to -1} f(x) = f(-1) \] ### Conclusion All three criteria for continuity at \( x = -1 \) are satisfied: 1. \( f(x) \) is defined at \( x = -1 \). 2. The limit of \( f(x) \) as \( x \) approaches \( -1 \) exists. 3. The limit of \( f(x) \) as \( x \) approaches \( -1 \) is equal to \( f(-1) \). Therefore, the function \( f(x) = \frac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \). |
True |
| llama3.1-70B-Instruct | To determine if the function f(x) = (4x+5)/(9-3x) is continuous at x = -1, we need to check if the following conditions are met: 1. The function is defined at x = -1. 2. The limit of the function as x approaches -1 exists. 3. The limit of the function as x approaches -1 is equal to the value of the function at x = -1. Let's check these conditions: 1. Is the function defined at x = -1? Substituting x = -1 into the function, we get: f(-1) = (4(-1)+5)/(9-3(-1)) = (-4+5)/(9+3) = 1/12 So, the function is defined at x = -1. 2. Does the limit of the function as x approaches -1 exist? We can evaluate the limit by substituting x = -1 into the function: lim x→-1 (4x+5)/(9-3x) = (4(-1)+5)/(9-3(-1)) = (-4+5)/(9+3) = 1/12 The limit exists and is equal to 1/12. 3. Is the limit of the function as x approaches -1 equal to the value of the function at x = -1? We already found that the limit is equal to 1/12, and the value of the function at x = -1 is also 1/12. Therefore, the limit is equal to the value of the function at x = -1. Since all three conditions are met, we can conclude that the function f(x) = (4x+5)/(9-3x) is continuous at x = -1. |
True |
| llama3.1-405B-Instruct | To determine if the function f(x) = (4x+5)/(9-3x) is continuous at x = -1, we need to check if the following conditions are satisfied: 1. The function is defined at x = -1. 2. The limit of the function as x approaches -1 exists. 3. The limit of the function as x approaches -1 is equal to the value of the function at x = -1. Let's check these conditions: 1. The function is defined at x = -1: f(-1) = (4(-1) + 5) / (9 - 3(-1)) = (-4 + 5) / (9 + 3) = 1 / 12 So, the function is defined at x = -1. 2. The limit of the function as x approaches -1 exists: We can evaluate the limit by plugging in x = -1 into the function: lim (x → -1) (4x + 5) / (9 - 3x) = (4(-1) + 5) / (9 - 3(-1)) = (-4 + 5) / (9 + 3) = 1 / 12 So, the limit exists and is equal to 1/12. 3. The limit of the function as x approaches -1 is equal to the value of the function at x = -1: We already found that the limit is 1/12, and we also found that f(-1) = 1/12. Therefore, the limit is equal to the value of the function at x = -1. Since all three conditions are satisfied, we can conclude that the function f(x) = (4x+5)/(9-3x) is continuous at x = -1. |
True |
| DeepSeek-Coder-V2 | To determine if the function \( f(x) = \frac{4x+5}{9-3x} \) is continuous at \( x = -1 \), we need to check if the following conditions are met: 1. \( f(-1) \) is defined. 2. \(\lim_{x \to -1} f(x)\) exists. 3. \(\lim_{x \to -1} f(x) = f(-1)\). Let's evaluate each condition step by step. ### Step 1: Check if \( f(-1) \) is defined Substitute \( x = -1 \) into the function \( f(x) \): \[ f(-1) = \frac{4(-1) + 5}{9 - 3(-1)} \] Calculate the numerator and the denominator separately: \[ 4(-1) + 5 = -4 + 5 = 1 \] \[ 9 - 3(-1) = 9 + 3 = 12 \] So, \[ f(-1) = \frac{1}{12} \] Thus, \( f(-1) \) is defined. ### Step 2: Check if \(\lim_{x \to -1} f(x)\) exists To find \(\lim_{x \to -1} f(x)\), we need to evaluate the limit from both sides (left-hand limit and right-hand limit) and see if they are equal. #### Left-hand limit: As \( x \) approaches \(-1\) from the left (\( x \to -1^- \)): \[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{4x + 5}{9 - 3x} \] Substitute \( x = -1 \): \[ \lim_{x \to -1^-} \frac{4x + 5}{9 - 3x} = \frac{1}{12} \] #### Right-hand limit: As \( x \) approaches \(-1\) from the right (\( x \to -1^+ \)): \[ \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{4x + 5}{9 - 3x} \] Substitute \( x = -1 \): \[ \lim_{x \to -1^+} \frac{4x + 5}{9 - 3x} = \frac{1}{12} \] Since both the left-hand limit and the right-hand limit are equal to \(\frac{1}{12}\), the limit exists and is: \[ \lim_{x \to -1} f(x) = \frac{1}{12} \] ### Step 3: Check if \(\lim_{x \to -1} f(x) = f(-1)\) From Step 1, we found that \( f(-1) = \frac{1}{12} \). From Step 2, we found that \(\lim_{x \to -1} f(x) = \frac{1}{12}\). Since \(\lim_{x \to -1} f(x) = f(-1)\), the function \( f(x) \) is continuous at \( x = -1 \). Therefore, the function \( f(x) = \frac{4x+5}{9-3x} \) is continuous at \( x = -1 \). |
True |
| Qwen2-72B-Instruct | To determine if the function \(f(x) = \frac{4x+5}{9-3x}\) is continuous at \(x = -1\), we need to check three conditions for continuity at that point: 1. \(f(-1)\) exists. 2. \(\lim_{x \to -1} f(x)\) exists. 3. \(\lim_{x \to -1} f(x) = f(-1)\). Let's check these conditions: 1. **Existence of \(f(-1)\):** Substitute \(x = -1\) into \(f(x)\): \[ f(-1) = \frac{4(-1)+5}{9-3(-1)} = \frac{-4+5}{9+3} = \frac{1}{12} \] So, \(f(-1)\) exists and is equal to \(\frac{1}{12}\). 2. **Existence of \(\lim_{x \to -1} f(x)\):** To find this limit, we can directly substitute \(x = -1\) into the function, as long as the denominator is not zero at \(x = -1\). Since we already evaluated \(f(-1)\), and the denominator was not zero, we can say: \[ \lim_{x \to -1} f(x) = \frac{1}{12} \] So, the limit exists. 3. **Equality of \(\lim_{x \to -1} f(x)\) and \(f(-1)\):** Since both \(\lim_{x \to -1} f(x)\) and \(f(-1)\) are equal to \(\frac{1}{12}\), they are equal. Since all three conditions are satisfied, we can conclude that the function \(f(x) = \frac{4x+5}{9-3x}\) is continuous at \(x = -1\). |
True |
| baichuan4 | To determine if the function \( f(x) = \frac{4x+5}{9-3x} \) is continuous at \( x = -1 \), we need to check if the function is defined at \( x = -1 \) and if its limit as \( x \) approaches \( -1 \) exists and is equal to \( f(-1) \). First, let's see if the function is defined at \( x = -1 \): \( f(-1) = \frac{4(-1)+5}{9-3(-1)} = \frac{-4+5}{9+3} = \frac{1}{12} \) The function is defined at \( x = -1 \) because it gives a finite value. Next, let's find the limit of \( f(x) \) as \( x \) approaches \( -1 \) from both sides: \( \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \frac{4x+5}{9-3x} \) \( \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{4x+5}{9-3x} \) Since the function is rational and the denominator does not become zero at \( x = -1 \), the limit from both sides will be the same: \( \lim_{x \to -1} f(x) = \frac{4(-1)+5}{9-3(-1)} = \frac{1}{12} \) Now, we compare the limit with the value of the function at \( x = -1 \): \( \lim_{x \to -1} f(x) = f(-1) = \frac{1}{12} \) Since the limit of \( f(x) \) as \( x \) approaches \( -1 \) is equal to the value of \( f(-1) \), we can conclude that the function \( f(x) \) is continuous at \( x = -1 \). |
True |
| doubao | To determine if the function is continuous at a particular point, we need to check if the limit of the function as x approaches that point exists and is equal to the value of the function at that point. First find the limit as x approaches -1: lim(x→-1) (4x+5)/(9-3x) = (4*(-1)+5)/(9-3*(-1)) = 1/12. Now find f(-1) = (4*(-1)+5)/(9-3*(-1)) = 1/12. Since the limit as x approaches -1 is equal to the value of the function at x=-1, the function is continuous at x=-1. |
True |
| minimax | To determine if the function f(x) = (4x + 5) / (9 - 3x) is continuous at x = -1, we need to check three conditions: 1. f(-1) is defined. 2. The limit of f(x) as x approaches -1 exists. 3. The limit of f(x) as x approaches -1 is equal to f(-1). Let's check each condition: 1. f(-1) is defined: To check if f(-1) is defined, we substitute x = -1 into the function: f(-1) = (4(-1) + 5) / (9 - 3(-1)) = (-4 + 5) / (9 + 3) = 1 / 12 Since the result is a real number, f(-1) is defined. 2. The limit of f(x) as x approaches -1 exists: We need to check if the limit of f(x) as x approaches -1 from both the left and the right is the same. Since the denominator 9 - 3x is a linear function and does not equal zero for x = -1, there are no issues with division by zero. Therefore, the limit exists and is equal to f(-1). 3. The limit of f(x) as x approaches -1 is equal to f(-1): We have already calculated f(-1) as 1/12. Since the function is a rational function and the denominator does not become zero at x = -1, the limit as x approaches -1 is equal to the value of the function at x = -1. Therefore, since all three conditions are satisfied, the function f(x) = (4x + 5) / (9 - 3x) is continuous at x = -1. |
True |
| kimi | To determine if the function \( f(x) = \frac{4x+5}{9-3x} \) is continuous at \( x = -1 \), we need to check two conditions: 1. The function must be defined at \( x = -1 \). 2. The limit of the function as \( x \) approaches \( -1 \) must exist and be equal to the value of the function at \( x = -1 \). Let's check these conditions: 1. **Is the function defined at \( x = -1 \)?** The function \( f(x) = \frac{4x+5}{9-3x} \) is defined for all \( x \) except where the denominator is zero. To find out where the denominator is zero, we set it equal to zero and solve for \( x \): \[ 9 - 3x = 0 \implies x = 3 \] Since the denominator is zero when \( x = 3 \), the function is not defined at \( x = 3 \). However, this does not affect the continuity at \( x = -1 \) because \( x = -1 \) is not equal to \( x = 3 \). So the function is defined at \( x = -1 \). 2. **Does the limit exist as \( x \) approaches \( -1 \)?** We calculate the limit of the function as \( x \) approaches \( -1 \): \[ \lim_{x \to -1} \frac{4x+5}{9-3x} \] Substitute \( x = -1 \) into the function: \[ \lim_{x \to -1} \frac{4(-1)+5}{9-3(-1)} = \frac{-4+5}{9+3} = \frac{1}{12} \] So the limit exists and is equal to \( \frac{1}{12} \). 3. **Is the limit equal to the value of the function at \( x = -1 \)?** Now we evaluate the function at \( x = -1 \): \[ f(-1) = \frac{4(-1)+5}{9-3(-1)} = \frac{-4+5}{9+3} = \frac{1}{12} \] The value of the function at \( x = -1 \) is also \( \frac{1}{12} \). Since both conditions are satisfied, the function \( f(x) = \frac{4x+5}{9-3x} \) is continuous at \( x = -1 \). |
True |
| wenxinyiyan | We need to check if the function $f(x) = \frac{4x + 5}{9 - 3x}$ is continuous at $x = -1$. First, let's find $f(-1)$: $f(-1) = \frac{4(-1) + 5}{9 - 3(-1)} = \frac{-4 + 5}{9 + 3} = \frac{1}{12}$ Since $f(-1)$ exists and is equal to $\frac{1}{12}$, we need to check if the limit of $f(x)$ as $x$ approaches $-1$ exists and is equal to $f(-1)$. $\lim_{{x \to -1}} f(x) = \lim_{{x \to -1}} \frac{4x + 5}{9 - 3x}$ To find the limit, we can directly substitute $x = -1$ into the function since it is a rational function and the denominator does not equal zero at $x = -1$. $\lim_{{x \to -1}} f(x) = \frac{4(-1) + 5}{9 - 3(-1)} = \frac{-4 + 5}{9 + 3} = \frac{1}{12}$ Since the limit of $f(x)$ as $x$ approaches $-1$ exists and is equal to $f(-1)$, we can conclude that the function $f(x) = \frac{4x + 5}{9 - 3x}$ is continuous at $x = -1$. |
True |
| zhipu | To determine if the function \( f(x) = \frac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \), we need to check three conditions: 1. The function \( f(x) \) is defined at \( x = -1 \). 2. The limit of \( f(x) \) as \( x \) approaches \( -1 \) exists. 3. The value of the function at \( x = -1 \) is equal to the limit of the function as \( x \) approaches \( -1 \). ### Step 1: Check if \( f(x) \) is defined at \( x = -1 \) To check if \( f(x) \) is defined at \( x = -1 \), we substitute \( x = -1 \) into the function: \[ f(-1) = \frac{4(-1) + 5}{9 - 3(-1)} = \frac{-4 + 5}{9 + 3} = \frac{1}{12} \] Since \( f(-1) = \frac{1}{12} \) is a finite number, the function is defined at \( x = -1 \). ### Step 2: Check if the limit of \( f(x) \) as \( x \) approaches \( -1 \) exists To find the limit of \( f(x) \) as \( x \) approaches \( -1 \), we substitute \( x = -1 \) into the function directly, as the function is a rational function and is continuous wherever it is defined: \[ \lim_{{x \to -1}} f(x) = \lim_{{x \to -1}} \frac{4x + 5}{9 - 3x} = \frac{4(-1) + 5}{9 - 3(-1)} = \frac{-4 + 5}{9 + 3} = \frac{1}{12} \] ### Step 3: Compare the value of the function at \( x = -1 \) with the limit From Step 1, we have \( f(-1) = \frac{1}{12} \), and from Step 2, we have \( \lim_{{x \to -1}} f(x) = \frac{1}{12} \). Since \( f(-1) = \lim_{{x \to -1}} f(x) \), the function \( f(x) \) is continuous at \( x = -1 \). ### Conclusion The function \( f(x) = \frac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \). |
True |
| gpto1 | To determine if the function \( f(x) = \dfrac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \), we need to check the following three conditions for continuity at that point: 1. **The function is defined at \( x = -1 \).** 2. **The limit of the function exists as \( x \) approaches \( -1 \).** 3. **The limit of the function as \( x \) approaches \( -1 \) equals the function's value at \( x = -1 \).** Let's examine each of these steps in detail. --- ### **1. The function is defined at \( x = -1 \).** First, evaluate \( f(-1) \): \[ \begin{align*} f(-1) &= \dfrac{4(-1) + 5}{9 - 3(-1)} \\ &= \dfrac{-4 + 5}{9 + 3} \\ &= \dfrac{1}{12} \end{align*} \] Since \( f(-1) = \dfrac{1}{12} \) is a real number, the function is defined at \( x = -1 \). --- ### **2. The limit of the function exists as \( x \) approaches \( -1 \).** Compute the limit of \( f(x) \) as \( x \) approaches \( -1 \): \[ \lim_{x \to -1} f(x) = \lim_{x \to -1} \dfrac{4x + 5}{9 - 3x} \] Since the denominator \( 9 - 3x \) does not equal zero at \( x = -1 \) (as calculated earlier, it equals \( 12 \)), and both the numerator and the denominator are continuous functions (polynomials are continuous everywhere), the function \( f(x) \) is continuous wherever it is defined, except possibly where the denominator is zero. Therefore, the limit exists and is equal to the function evaluated at \( x = -1 \): \[ \lim_{x \to -1} f(x) = f(-1) = \dfrac{1}{12} \] --- ### **3. The limit equals the function's value at \( x = -1 \).** From the previous steps: \[ \lim_{x \to -1} f(x) = \dfrac{1}{12} \quad \text{and} \quad f(-1) = \dfrac{1}{12} \] Since the limit of \( f(x) \) as \( x \) approaches \( -1 \) equals the function's value at that point, the third condition for continuity is satisfied. --- ### **Conclusion** All three conditions are met: 1. \( f(-1) \) is defined. 2. \( \lim_{x \to -1} f(x) \) exists. 3. \( \lim_{x \to -1} f(x) = f(-1) \). **Therefore, the function \( f(x) = \dfrac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \).** |
True |
| deepseekv3 | To determine whether the function \( f(x) = \frac{4x + 5}{9 - 3x} \) is continuous at \( x = -1 \), we need to check the following conditions: 1. **The function is defined at \( x = -1 \).** This means the denominator must not be zero at \( x = -1 \). 2. **The limit of the function as \( x \) approaches \(-1 \) exists.** 3. **The value of the function at \( x = -1 \) equals the limit as \( x \) approaches \(-1\).** Let’s evaluate each condition: --- ### Step 1: Check if the function is defined at \( x = -1 \) Substitute \( x = -1 \) into the denominator: \[ 9 - 3(-1) = 9 + 3 = 12 \neq 0. \] Since the denominator is not zero, the function is defined at \( x = -1 \). --- ### Step 2: Compute the limit as \( x \) approaches \(-1 \) The function \( f(x) = \frac{4x + 5}{9 - 3x} \) is a rational function, and rational functions are continuous everywhere except where the denominator is zero. Since the denominator is not zero at \( x = -1 \), the limit exists and is equal to the value of the function at \( x = -1 \). --- ### Step 3: Evaluate the function at \( x = -1 \) Substitute \( x = -1 \) into the function: \[ f(-1) = \frac{4(-1) + 5}{9 - 3(-1)} = \frac{-4 + 5}{9 + 3} = \frac{1}{12}. \] --- ### Conclusion Since the function is defined at \( x = -1 \), the limit exists, and the value of the function equals the limit, the function \( f(x) \) is **continuous at \( x = -1 \)**. |
True |