Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.
