Answer :
To solve the problem step by step, let's analyze the function [tex]\( y = x^{\frac{6}{7}} (x^2 - 4) \)[/tex].
### Step 1: Domain Endpoints
The function [tex]\( y = x^{\frac{6}{7}} (x^2 - 4) \)[/tex] is defined for all real values of [tex]\( x \)[/tex]. There are no restrictions on the domain because the exponent [tex]\( \frac{6}{7} \)[/tex] is positive and non-zero for any real [tex]\( x \)[/tex]. Hence, the function does not have any endpoints within the real number set.
Answer:
B. There are no domain endpoints.
### Step 2: Local Maxima and Minima
To find local maxima and minima, we need to follow several steps:
#### Taking the First Derivative
First, we find the first derivative [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ y = x^{\frac{6}{7}} (x^2 - 4) \][/tex]
Let's apply the product rule:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}\left(x^{\frac{6}{7}}\right) \cdot (x^2 - 4) + x^{\frac{6}{7}} \cdot \frac{d}{dx}(x^2 - 4) \][/tex]
Using the power rule,
[tex]\[ \frac{d}{dx}\left(x^{\frac{6}{7}}\right) = \frac{6}{7} x^{\frac{-1}{7}} \][/tex]
and
[tex]\[ \frac{d}{dx}(x^2 - 4) = 2x \][/tex]
So the first derivative is:
[tex]\[ \frac{dy}{dx} = \frac{6}{7} x^{\frac{-1}{7}} (x^2 - 4) + x^{\frac{6}{7}} 2x \][/tex]
[tex]\[ = \frac{6}{7} x^{\frac{-1}{7}} (x^2 - 4) + 2x^{\frac{13}{7}} \][/tex]
Set the derivative to zero to find critical points:
[tex]\[ \frac{6}{7} x^{\frac{-1}{7}} (x^2 - 4) + 2x^{\frac{13}{7}} = 0 \][/tex]
Combine like terms:
[tex]\[ \frac{6}{7} (x^2 - 4) + 2x^2 = 0 \][/tex]
Multiply through by [tex]\( x^{\frac{1}{7}} \)[/tex]:
[tex]\[ \frac{6}{7} (x^2 - 4)x^{-\frac{1}{7}} + 2x^{2+\frac{6}{7}} = 0 \][/tex]
Simplify:
[tex]\[ \frac{6}{7} x^{-\frac{1}{7}} (x^2 - 4) + 2x^{\frac{20}{7}} = 0 \][/tex]
[tex]\[ \frac{6}{7} (x^2 - 4) + 2x^{\frac{21}{7}} = 0 \][/tex]
Separate terms:
[tex]\[ \frac{6}{7} (x^2 - 4) + \frac{14x^3}{7} = 0 \][/tex]
Equate coefficients:
[tex]\[ 6(x^2 - 4) + 14x^3 = 0 \][/tex]
[tex]\[ 6x^2 - 24 + 14x^3 = 0 \][/tex]
Factor:
[tex]\[ 2(x^2 - 4) + x^3 = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x^3 + 6x - 8x = 0 \][/tex]
[tex]\[ x(2x^2 + 6 - 8) = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
For higher order terms:
[tex]\[ x^3 + \frac{14}{7}(x^2 - 4) = 0 \][/tex]
Solving cubic eliminates real roots:
[tex]\[ x = +1.095, -1.095\][/tex]
### Step 3: Local Maxima and Minima
To determine whether these critical points are maxima or minima, use the second derivative test:
[tex]\[ \frac{d^2y}{dx^2} \][/tex]
Local Maxima: No local maximum point exists.
Local Minima: The critical points [tex]\( +1.095 \)[/tex] and [tex]\( -1.095 \)[/tex] and their corresponding y-values are:
Plug into the original function to find [tex]\( y \)[/tex]:
[tex]\[ y(-1.095) = (-1.095)^{6/7} \times (-1.095^2 - 4) = -3.028 \][/tex]
Similarlity opposite points:
[tex]\[ y(1.095) = (-1.095)^{6/7} \times (-1.095^2 - 4) = -3.028 \][/tex]
### Step 4: Absolute Maxima
Evaluate the value's if maxima:
[tex]\[ y'(0) \][/tex]
Values not yielded thus,
Answer:
A. (-1.095, -3.028),(1.095, -3.028)
### Step 1: Domain Endpoints
The function [tex]\( y = x^{\frac{6}{7}} (x^2 - 4) \)[/tex] is defined for all real values of [tex]\( x \)[/tex]. There are no restrictions on the domain because the exponent [tex]\( \frac{6}{7} \)[/tex] is positive and non-zero for any real [tex]\( x \)[/tex]. Hence, the function does not have any endpoints within the real number set.
Answer:
B. There are no domain endpoints.
### Step 2: Local Maxima and Minima
To find local maxima and minima, we need to follow several steps:
#### Taking the First Derivative
First, we find the first derivative [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ y = x^{\frac{6}{7}} (x^2 - 4) \][/tex]
Let's apply the product rule:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}\left(x^{\frac{6}{7}}\right) \cdot (x^2 - 4) + x^{\frac{6}{7}} \cdot \frac{d}{dx}(x^2 - 4) \][/tex]
Using the power rule,
[tex]\[ \frac{d}{dx}\left(x^{\frac{6}{7}}\right) = \frac{6}{7} x^{\frac{-1}{7}} \][/tex]
and
[tex]\[ \frac{d}{dx}(x^2 - 4) = 2x \][/tex]
So the first derivative is:
[tex]\[ \frac{dy}{dx} = \frac{6}{7} x^{\frac{-1}{7}} (x^2 - 4) + x^{\frac{6}{7}} 2x \][/tex]
[tex]\[ = \frac{6}{7} x^{\frac{-1}{7}} (x^2 - 4) + 2x^{\frac{13}{7}} \][/tex]
Set the derivative to zero to find critical points:
[tex]\[ \frac{6}{7} x^{\frac{-1}{7}} (x^2 - 4) + 2x^{\frac{13}{7}} = 0 \][/tex]
Combine like terms:
[tex]\[ \frac{6}{7} (x^2 - 4) + 2x^2 = 0 \][/tex]
Multiply through by [tex]\( x^{\frac{1}{7}} \)[/tex]:
[tex]\[ \frac{6}{7} (x^2 - 4)x^{-\frac{1}{7}} + 2x^{2+\frac{6}{7}} = 0 \][/tex]
Simplify:
[tex]\[ \frac{6}{7} x^{-\frac{1}{7}} (x^2 - 4) + 2x^{\frac{20}{7}} = 0 \][/tex]
[tex]\[ \frac{6}{7} (x^2 - 4) + 2x^{\frac{21}{7}} = 0 \][/tex]
Separate terms:
[tex]\[ \frac{6}{7} (x^2 - 4) + \frac{14x^3}{7} = 0 \][/tex]
Equate coefficients:
[tex]\[ 6(x^2 - 4) + 14x^3 = 0 \][/tex]
[tex]\[ 6x^2 - 24 + 14x^3 = 0 \][/tex]
Factor:
[tex]\[ 2(x^2 - 4) + x^3 = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x^3 + 6x - 8x = 0 \][/tex]
[tex]\[ x(2x^2 + 6 - 8) = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
For higher order terms:
[tex]\[ x^3 + \frac{14}{7}(x^2 - 4) = 0 \][/tex]
Solving cubic eliminates real roots:
[tex]\[ x = +1.095, -1.095\][/tex]
### Step 3: Local Maxima and Minima
To determine whether these critical points are maxima or minima, use the second derivative test:
[tex]\[ \frac{d^2y}{dx^2} \][/tex]
Local Maxima: No local maximum point exists.
Local Minima: The critical points [tex]\( +1.095 \)[/tex] and [tex]\( -1.095 \)[/tex] and their corresponding y-values are:
Plug into the original function to find [tex]\( y \)[/tex]:
[tex]\[ y(-1.095) = (-1.095)^{6/7} \times (-1.095^2 - 4) = -3.028 \][/tex]
Similarlity opposite points:
[tex]\[ y(1.095) = (-1.095)^{6/7} \times (-1.095^2 - 4) = -3.028 \][/tex]
### Step 4: Absolute Maxima
Evaluate the value's if maxima:
[tex]\[ y'(0) \][/tex]
Values not yielded thus,
Answer:
A. (-1.095, -3.028),(1.095, -3.028)
