If f and g are twice differentiable functions and $f(p)=3$, $f^{\prime}(p)=-2$, $g(p)=-1$, $g^{\prime}(p)=4$, then: $lim_{x\rightarrow p}\frac{g(x)f(p)-g(p)f(x)}{x-p}=$

Q: If f and g are twice differentiable functions and $f(p)=3$, $f^{\prime}(p)=-2$, $g(p)=-1$, $g^{\prime}(p)=4$, then: $lim_{x\rightarrow p}\frac{g(x)f(p)-g(p)f(x)}{x-p}=$

To solve the problem, we need to evaluate the limit: \[ \lim_{x \rightarrow p} \frac{g(x)f(p) - g(p)f(x)}{x-p} \] Given the values: - $ f(p) = 3 $ - $ f'(p) = -2 $ - $ g(p) = -1 $ - $ g'(p) = 4 $ We can rewrite the expression inside the limit as: \[ \frac{g(x)f(p) - g(p)f(x)}{x-p} = \frac{g(x) \cdot 3 - (-1) \cdot f(x)}{x-p} = \frac{3g(x) + f(x)}{x-p} \] This expression is in the form of a difference quotient, which suggests the use of L'Hôpital's Rule, since both the numerator and the denominator approach zero as $ x \rightarrow p $. Applying L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to $ x $: 1. Differentiate the numerator: $ \frac{d}{dx}[3g(x) + f(x)] = 3g'(x) + f'(x) $ 2. Differentiate the denominator: $ \frac{d}{dx}[x-p] = 1 $ Now, evaluate the derivatives at $ x = p $: - $ g'(p) = 4 $ - $ f'(p) = -2 $ Substitute these values into the differentiated expression: \[ 3g'(p) + f'(p) = 3 \cdot 4 + (-2) = 12 - 2 = 10 \] Therefore, the limit evaluates to: \[ \lim_{x \rightarrow p} \frac{3g(x) + f(x)}{x-p} = 10 \] Thus, the correct answer is Option 4: 10. **Explanation for Incorrect Options:** 1. **Option 1 (5):** This option is incorrect because the calculation of the limit using L'Hôpital's Rule results in 10, not 5. The derivatives of the functions at $ p $ do not sum to 5. 2. **Option 2 (-5):** This option is incorrect because the correct application of L'Hôpital's Rule yields a positive result of 10, not a negative value like -5. 3. **Option 3 (-10):** This option is incorrect because the calculation shows that the limit is positive 10, not negative 10. The signs of the derivatives and their coefficients do not lead to a negative result. In conclusion, the correct answer is indeed Option 4: 10, as supported by the application of L'Hôpital's Rule and the given derivatives.

Question

Accepted Answer

To solve the problem, we need to evaluate the limit:

$$
\lim_{x \rightarrow p} \frac{g(x)f(p) - g(p)f(x)}{x-p}
$$

Given the values:
- $ f(p) = 3 $
- $ f'(p) = -2 $
- $ g(p) = -1 $
- $ g'(p) = 4 $

We can rewrite the expression inside the limit as:

$$
\frac{g(x)f(p) - g(p)f(x)}{x-p} = \frac{g(x) \cdot 3 - (-1) \cdot f(x)}{x-p} = \frac{3g(x) + f(x)}{x-p}
$$

This expression is in the form of a difference quotient, which suggests the use of L'Hôpital's Rule, since both the numerator and the denominator approach zero as $ x \rightarrow p $.

Applying L'Hôpital's Rule, we differentiate the numerator and the denominator with respect to $ x $:

1. Differentiate the numerator: $ \frac{d}{dx}[3g(x) + f(x)] = 3g'(x) + f'(x) $
2. Differentiate the denominator: $ \frac{d}{dx}[x-p] = 1 $

Now, evaluate the derivatives at $ x = p $:

- $ g'(p) = 4 $
- $ f'(p) = -2 $

Substitute these values into the differentiated expression:

$$
3g'(p) + f'(p) = 3 \cdot 4 + (-2) = 12 - 2 = 10
$$

Therefore, the limit evaluates to:

$$
\lim_{x \rightarrow p} \frac{3g(x) + f(x)}{x-p} = 10
$$

Thus, the correct answer is Option 4: 10.

**Explanation for Incorrect Options:**

1. **Option 1 (5):** This option is incorrect because the calculation of the limit using L'Hôpital's Rule results in 10, not 5. The derivatives of the functions at $ p $ do not sum to 5.

2. **Option 2 (-5):** This option is incorrect because the correct application of L'Hôpital's Rule yields a positive result of 10, not a negative value like -5.

3. **Option 3 (-10):** This option is incorrect because the calculation shows that the limit is positive 10, not negative 10. The signs of the derivatives and their coefficients do not lead to a negative result.

In conclusion, the correct answer is indeed Option 4: 10, as supported by the application of L'Hôpital's Rule and the given derivatives.