Answer:
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:
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.
2) After that, set this an equation then find the values for x.
b) The Second Derivative Test helps us to check the sign of given critical numbers.
Rewriting f'(x) factorizing:
Applying product Rule to find the 2nd Derivative, similarly to 1st derivative:
1) Setting this to zero, as an equation:
2) Now, let's define which is the inflection point, the domain is as a polynomial function:
Looking at the graph.
Plugging these inflection points in the original equation to get y coordinate:
We have as Inflection Points and their respective y coordinates (Converting to approximate decimal numbers)
Inflection Point and Local Minimum
Inflection Point and Saddle Point
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.
At
Local Minimum
(Saddle Point)
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.
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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Answer:
It's 21, you are correct!!
Step-by-step explanation:
f(x)
x y
0 −5
1 0
2 3
3 4
4 3
5 0
6 −5
Which function has the largest maximum?
Answers:
f(x)
g(x)
h(x)
All three functions have the same maximum value.
Answer:
The correct option is 1. The function f(x) has the largest maximum.
Step-by-step explanation:
The vertex form of a parabola is
Where, (h,k) is vertex.
The given functions is
Here, a=-1, h=-5 and k=3. Since the value of a is negative, therefore it is an downward parabola and vertex is the point of maxima.
Thus the maximum value of the function h(x) is 3.
The value of cosine function lies between -1 to 1.
Multiply 4 on each side.
Subtract 2 from each side.
Therefore the maximum value of the function g(x) is 2.
From the given table it is clear that the maximum value of the function f(x) is 4 at x=3.
Since the function f(x) has the largest maximum, therefore the correct option is 1.
Answer:
The probability that the student is in high school and has visited a foreign country is 0.18.
Step-by-step explanation:
We are given that in a survey of students, 60% were in high school and 40% were in middle school.
Of the high school students, 30% had visited a foreign country.
Let the Probability that students were in high school = P(H) = 60%
Probability that students were in middle school = P(M) = 40%
Also, let F = event that students had had visited a foreign country
So, Probability that high school students had visited a foreign country = P(F/H) = 30%
Now, probability that the student is in high school and has visited a foreign country is given by = Probability that students were in high school Probability that high school students had visited a foreign country
= P(H) P(F/H)
= 0.60 0.30
= 0.18 or 18%
x+6y=-3
Would you rather use subtraction or addition to solve the system? Explain
your reasoning. (Lesson 2-14)
It is more convenient to use addition to solve the system of equations as it eliminates one of the variables when the equations are added. The solution to the system of equations is x = -6 and y = 1/2.
To solve the given system of equations, it is generally more convenient to use addition rather than subtraction. This is because addition eliminates one of the variables when the equations are added together, making it easier to solve for the remaining variable. Let's see how:
Adding the two equations together eliminates the 'x' variable:
-x + 6y + x + 6y = 9 + (-3)
12y = 6
Now, we can solve for 'y' by dividing both sides by 12:
y = 6/12 = 1/2
Substituting the value of 'y' back into either of the original equations, we can solve for 'x':
-x + 6(1/2) = 9
-x + 3 = 9
-x = 6
x = -6
Therefore, the solution to the system of equations is x = -6 and y = 1/2.
Answer: True
Step-by-step explanation:
Let p= Students who do well in course do not skip class
q= Student who study hard do well in course
So p^q= Student who study hard and who do well in course do not skip class.
If p= true and q=true then p^q= true by discrete maths.
The argument is valid because the conclusion is logically derived from the provided premises. However, it is important to note that the validity of an argument does not guarantee the truth of its premises. The argument may be valid, but its premises could still be false.
The argument provided is valid.
The reasoning follows a valid logicalstructure, specifically a form of argument called a syllogism, where conclusions are drawn from two or more premises. Let's break it down:
"For students to do well in a discrete mathematics course, it is necessary that they study hard." This is a premise, stating that studying hard is a necessary condition for success in a discrete mathematics course.
"Students who do well in courses do not skip classes." This is another premise, suggesting that students who perform well in their courses do not miss classes.
"Students who study hard do well in courses." This is also a premise, indicating that diligent study leads to success in courses.
The conclusion drawn is: "Therefore students who do well in a discrete mathematics course do not skip class." This conclusion logically follows from the given premises. If we accept the truth of the premises, we must also accept the truth of the conclusion.
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