Get the information you need with the help of IDNLearn.com's extensive Q&A platform. Discover the information you need quickly and easily with our reliable and thorough Q&A platform.
Sagot :
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)
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for trusting IDNLearn.com with your questions. Visit us again for clear, concise, and accurate answers.
