Answer :
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]
