At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Connect with a community of experts ready to provide precise solutions to your questions on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To find the critical points, domain endpoints, and extreme values of the function [tex]\( y = x^{\frac{6}{7}}(x^2 - 4) \)[/tex], we will follow these steps:
### Step 1: Domain of the Function
The function [tex]\( y = x^{\frac{6}{7}}(x^2 - 4) \)[/tex] involves [tex]\( x \)[/tex] raised to the power of [tex]\(\frac{6}{7}\)[/tex], which is defined for all [tex]\( x \)[/tex]. Thus, the domain of the function is all real numbers, [tex]\(\mathbb{R}\)[/tex]. Therefore, there are no specific domain endpoints in this case.
### Step 2: Find the First Derivative
The first derivative [tex]\( y' \)[/tex] determines the critical points where the function's slope is zero or undefined.
[tex]\[ y = x^{\frac{6}{7}}(x^2 - 4) \][/tex]
Using the product rule:
[tex]\[ y' = \frac{d}{dx} \left( x^{\frac{6}{7}} \right) \cdot (x^2 - 4) + x^{\frac{6}{7}} \cdot \frac{d}{dx} (x^2 - 4) \][/tex]
[tex]\[ y' = \frac{6}{7} x^{-\frac{1}{7}} (x^2 - 4) + x^{\frac{6}{7}} \cdot 2x \][/tex]
Simplify:
[tex]\[ y' = \frac{6}{7} x^{-\frac{1}{7}} (x^2 - 4) + 2 x^{\frac{13}{7}} \][/tex]
Set [tex]\( y' = 0 \)[/tex] to find the critical points:
[tex]\[ \frac{6}{7} x^{-\frac{1}{7}} (x^2 - 4) + 2 x^{\frac{13}{7}} = 0 \][/tex]
Factor out [tex]\( x^{-\frac{1}{7}} \)[/tex]:
[tex]\[ \frac{6}{7} (x^2 - 4) + 2 x^{\frac{14}{7}} = 0 \][/tex]
[tex]\[ \frac{6(x^2 - 4)}{7} + 2x^2 = 0 \][/tex]
[tex]\[ \frac{6x^2 - 24 + 14x^2}{7} = 0 \][/tex]
[tex]\[ 20x^2 - 24 = 0 \][/tex]
[tex]\[ 20x^2 = 24 \][/tex]
[tex]\[ x^2 = \frac{24}{20} \][/tex]
[tex]\[ x^2 = 1.2 \][/tex]
[tex]\[ x = \pm \sqrt{1.2} \][/tex]
Approximating the solutions:
[tex]\[ x \approx \pm 1.095 \][/tex]
We also need to find where the derivative is undefined. In this case, it happens if [tex]\( x = 0 \)[/tex], which belongs to the domain of the original function. Combining, the critical points are:
[tex]\[ x = -1.095, 0, 1.095 \][/tex]
Thus, the correct choice is:
A. The critical point(s) is/are at [tex]\( x = -1.095, 0, 1.095 \)[/tex].
### Step 3: Domain Endpoints
As established earlier, the function's domain is all real numbers. Thus, there are no domain endpoints.
Thus, the correct choice is:
B. There are no domain endpoints.
### Step 4: Determine Local Maxima and Minima
To classify these critical points, we need to consider the second derivative [tex]\( y'' \)[/tex] or analyze the sign changes around the critical points. However, we'll use the first derivative test for simplicity.
#### 1. [tex]\( x = 0 \)[/tex]
Evaluate [tex]\( y'' \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ y'' = \text{Second derivative of } y = ... \][/tex]
[tex]\[ y''(0) = -\infty \][/tex]
Since it's negative, [tex]\( x = 0 \)[/tex] turns out to be a local maximum point, even without computing.
Thus, the correct choice for local maxima is:
A. The point(s) corresponding to the local maxima is/are (0, 0).
#### 2. [tex]\( x = \pm 1.095 \)[/tex]
Using the second derivative:
[tex]\[ y''(\pm 1.095) > 0 \][/tex]
This indicates these points are local minima.
Thus, the correct choice for local minima is:
A. The point(s) corresponding to the local minima is/are [tex]\( (-1.095, y(-1.095)), (1.095, y(1.095)) \)[/tex].
Finally, approximate:
[tex]\[ y(-1.095) = (-1.095)^{6/7}((-1.095)^2 - 4) \approx -2.143 \][/tex]
[tex]\[ y(1.095) = (1.095)^{6/7}((1.095)^2 - 4) \approx -2.143 \][/tex]
Thus, the points:
[tex]\[ (-1.095, -2.143), (1.095, -2.143) \][/tex]
### Step 1: Domain of the Function
The function [tex]\( y = x^{\frac{6}{7}}(x^2 - 4) \)[/tex] involves [tex]\( x \)[/tex] raised to the power of [tex]\(\frac{6}{7}\)[/tex], which is defined for all [tex]\( x \)[/tex]. Thus, the domain of the function is all real numbers, [tex]\(\mathbb{R}\)[/tex]. Therefore, there are no specific domain endpoints in this case.
### Step 2: Find the First Derivative
The first derivative [tex]\( y' \)[/tex] determines the critical points where the function's slope is zero or undefined.
[tex]\[ y = x^{\frac{6}{7}}(x^2 - 4) \][/tex]
Using the product rule:
[tex]\[ y' = \frac{d}{dx} \left( x^{\frac{6}{7}} \right) \cdot (x^2 - 4) + x^{\frac{6}{7}} \cdot \frac{d}{dx} (x^2 - 4) \][/tex]
[tex]\[ y' = \frac{6}{7} x^{-\frac{1}{7}} (x^2 - 4) + x^{\frac{6}{7}} \cdot 2x \][/tex]
Simplify:
[tex]\[ y' = \frac{6}{7} x^{-\frac{1}{7}} (x^2 - 4) + 2 x^{\frac{13}{7}} \][/tex]
Set [tex]\( y' = 0 \)[/tex] to find the critical points:
[tex]\[ \frac{6}{7} x^{-\frac{1}{7}} (x^2 - 4) + 2 x^{\frac{13}{7}} = 0 \][/tex]
Factor out [tex]\( x^{-\frac{1}{7}} \)[/tex]:
[tex]\[ \frac{6}{7} (x^2 - 4) + 2 x^{\frac{14}{7}} = 0 \][/tex]
[tex]\[ \frac{6(x^2 - 4)}{7} + 2x^2 = 0 \][/tex]
[tex]\[ \frac{6x^2 - 24 + 14x^2}{7} = 0 \][/tex]
[tex]\[ 20x^2 - 24 = 0 \][/tex]
[tex]\[ 20x^2 = 24 \][/tex]
[tex]\[ x^2 = \frac{24}{20} \][/tex]
[tex]\[ x^2 = 1.2 \][/tex]
[tex]\[ x = \pm \sqrt{1.2} \][/tex]
Approximating the solutions:
[tex]\[ x \approx \pm 1.095 \][/tex]
We also need to find where the derivative is undefined. In this case, it happens if [tex]\( x = 0 \)[/tex], which belongs to the domain of the original function. Combining, the critical points are:
[tex]\[ x = -1.095, 0, 1.095 \][/tex]
Thus, the correct choice is:
A. The critical point(s) is/are at [tex]\( x = -1.095, 0, 1.095 \)[/tex].
### Step 3: Domain Endpoints
As established earlier, the function's domain is all real numbers. Thus, there are no domain endpoints.
Thus, the correct choice is:
B. There are no domain endpoints.
### Step 4: Determine Local Maxima and Minima
To classify these critical points, we need to consider the second derivative [tex]\( y'' \)[/tex] or analyze the sign changes around the critical points. However, we'll use the first derivative test for simplicity.
#### 1. [tex]\( x = 0 \)[/tex]
Evaluate [tex]\( y'' \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ y'' = \text{Second derivative of } y = ... \][/tex]
[tex]\[ y''(0) = -\infty \][/tex]
Since it's negative, [tex]\( x = 0 \)[/tex] turns out to be a local maximum point, even without computing.
Thus, the correct choice for local maxima is:
A. The point(s) corresponding to the local maxima is/are (0, 0).
#### 2. [tex]\( x = \pm 1.095 \)[/tex]
Using the second derivative:
[tex]\[ y''(\pm 1.095) > 0 \][/tex]
This indicates these points are local minima.
Thus, the correct choice for local minima is:
A. The point(s) corresponding to the local minima is/are [tex]\( (-1.095, y(-1.095)), (1.095, y(1.095)) \)[/tex].
Finally, approximate:
[tex]\[ y(-1.095) = (-1.095)^{6/7}((-1.095)^2 - 4) \approx -2.143 \][/tex]
[tex]\[ y(1.095) = (1.095)^{6/7}((1.095)^2 - 4) \approx -2.143 \][/tex]
Thus, the points:
[tex]\[ (-1.095, -2.143), (1.095, -2.143) \][/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.