Find the critical numbers of the function f(x) = x6(x − 2)5.x = (smallest value)x = x = (largest value)(b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers?At x = the function has ---Select--- [a local minimum, a local maximum, or neither a minimum nor a maximum].(c) What does the First Derivative Test tell you that the Second Derivative test does not? (Enter your answers from smallest to largest x value.)At x = the function has ---Select--- [a local minimum, a local maximum, or neither a minimum nor a maximum].At x = the function has ---Select--- [a local minimum, a local maximum, or neither a minimum nor a maximum].

Answers

Answer 1

Answer:

Answer:

$a) x=0, x=(12)/(11), x=2 \:$ b) The 2nd Derivative test shows us the change of sign and concavity at some point. c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

Step-by-step explanation:

a) To find the critical numbers, or critical points of:

$f(x)=x^(6)(x-2)^(5)$

1) The procedure is to calculate the 1st derivative of this function. Notice that in this case, we'll apply the Product Rule since there is a product of two functions.

$f(x)=x^(6)(x-2)^(5)\Rightarrow f'(x)=(f*g)'(x)\n=f'g+fg'\Rightarrow (fg)'(x)=6x^(5)(x-2)^(5)+5x^(6)(x-2)^(4) \Rightarrow 6x^(5)(x-2)^(5)+5x^(6)(x-2)^(4)=0\nf'(x)=6x^(5)(x-2)^(5)+5x^(6)(x-2)^(4)$

2) After that, set this an equation then find the values for x.

$x^(5)(x-2)^(4)[6(x-2)+5x]=0\Rightarrow x^(5)(x-2)^(4)[11x-12]=0\Rightarrow x_(1)=0\n(x-2)^(4)=0\Rightarrow \sqrt[4]{(x-2)}=\sqrt[4]{0}\Rightarrow x-2=0\Rightarrow x_(2)=2\n(11x-12)=0\Rightarrow x_(3)=(12)/(11)$

$x=0\:(smallest\:value)\:x_(3)=(12)/(11)\:x=2 (largest value)$

b) The Second Derivative Test helps us to check the sign of given critical numbers.

Rewriting f'(x) factorizing:

$f'(x)=(11x-12)(x-2)^4x^(5)$

Applying product Rule to find the 2nd Derivative, similarly to 1st derivative:

$f''(x)>0 \Rightarrow Concavity\: Up\n\nf''(x)<0\Rightarrow Concavity\:down$

$f''(x)=11\left(x-2\right)^4x^5+4\left(x-2\right)^3x^5\left(11x-12\right)+5\left(x-2\right)^4x^4\left(11x-12\right)\nf''(x)=10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)$

1) Setting this to zero, as an equation:

$10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0\n\n$

$10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0\n(x-2)^(3)=0 \Rightarrow x_1=2\nx^(4)=0 \therefore x_2=0\n11x^(2)-24x+12=0 \Rightarrow x_3=(12+2√(3))/(11)\:,x_4=(12-2√(3))/(11)\cong 0.78$

2) Now, let's define which is the inflection point, the domain is as a polynomial function:

$D=(-\infty<x<\infty)$

Looking at the graph.

Plugging these inflection points in the original equation $f(x)=x^(6)(x-2)^(5)$ to get y coordinate:

We have as Inflection Points and their respective y coordinates (Converting to approximate decimal numbers)

$(1.09,-1.05)$ Inflection Point and Local Minimum

$(2,0)$ Inflection Point and Saddle Point

$(0,0)$ Inflection Point Local Maximum

(Check the graph)

c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

$x=(12)/(11)\cong1.09$ Local Minimum

$At\:x=0,\:Local \:Maximum$

$At\:x=2, \:neither\:a\:minimum\:nor\:a\:maximum$ (Saddle Point)

Answer 2

Answer:

Final answer:

To find the critical numbers of the function f(x) = x^6(x - 2)^5, we need to set the first derivative equal to zero and solve for x. The Second Derivative Test tells us the behavior of the function at the critical numbers, while the First Derivative Test tells us the behavior of the function based on the sign change of the derivative at the critical numbers.

Explanation:

The critical numbers of the function f(x) = x^6(x - 2)^5 can be found by taking the first and second derivatives of the function. The first derivative is f'(x) = 6x^5(x - 2)^5 + 5x^6(x - 2)^4 and the second derivative is f''(x) = 30x^4(x - 2)^5 + 20x^5(x - 2)^4.

To find the critical numbers, we need to set the first derivative equal to zero and solve for x: 6x^5(x - 2)^5 + 5x^6(x - 2)^4 = 0. We can solve this equation using factoring or by using the Zero Product Property. Once we find the values of x that make the first derivative zero, we can evaluate the second derivative at those values to determine the behavior of the function at those critical numbers.

The Second Derivative Test tells us that if the second derivative is positive at a critical number, then the function has a local minimum at that point. If the second derivative is negative at a critical number, then the function has a local maximum at that point. If the second derivative is zero, the test is inconclusive and we need to use additional information to determine the behavior of the function. The First Derivative Test tells us that if the derivative changes sign from negative to positive at a critical number, then the function has a local minimum at that point. If the derivative changes sign from positive to negative at a critical number, then the function has a local maximum at that point.

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Answers

Answer:

(1,1) and (2,4) are the solutions to the system of equations.

Step-by-step explanation:

The two equations are:

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$y=3x-2$ ---------2

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Step-by-step explanation:

The system of equations given equation is

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Substituting 1 into equation 2, it becomes

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Answers

The value of the car in 2020 is $5532.53

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Given that, You purchase a car in 2010 for $25,000. The value of the car decreases by 14% annually.

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Hence, the value of the car in 2020 is $5532.53

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