Question 1

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; saddle point: 
B)  ; relative maximum value: 
C)  ; relative minimum value:

Accepted Answer

We need to find critical points first. To find these points we need to find the derivative of the given function, and set it equal to zero, then solve for $x$.

$f(x) = x^3 - 3x$
$f'(x) = 3x^2 - 3$

$3x^2 - 3 = 0 \Rightarrow x = \pm 1$

Therefore, there are two critical points: $(-1, 4)$ and $(1,-4)$.

Now we use the second derivative test to classify the nature of each point.

$f''(x) = 6x$, so $f''(-1) = -6$ and $f''(1) = 6$.

At $x=-1$, $f''(-1) < 0$, so it is a saddle point.

At $x=1$, $f''(1) > 0$, so it is a relative minimum value.

Therefore the answer is C, and the explanation shows that the function has only one relative minimum point at $(1,-4)$.

Question 2

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  and  ; saddle point:  ; relative maximum value:  ​
B)  and  ; relative minimum:  ; relative maximum value: 
C)  and  ; saddle point:  ; relative minimum value:  ​
D) there are no critical points

Accepted Answer

To find the critical points, we need to find the points where the first derivative is zero or undefined. Here, no function is given, so we cannot find the first derivative. Therefore, there are no critical points, except if the function is a constant, in which case every point is a critical point. However, we cannot determine the nature of these critical points or relative extrema without knowledge of the original function.

Question 3

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  is the critical point, it is impossible to determine the relative extrema of the function
B)  is the point of minimum, 9 is the relative minimum
C)  is the critical point, the function has neither a relative maximum nor a relative minimum at this point
D)  is the point of maximum, 9 is the relative maximum
E) no critical points

Accepted Answer

The answer of Find the critical point(s) of the function....

Question 4

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  and  ; relative maximum value:  ; relative minimum value: 
B)  and  ; saddle point:  ; relative minimum value: 
C)  ; saddle point: 
D)  ; relative maximum value: 
E) there are no critical points

Accepted Answer

The critical points are where the derivative is equal to zero or undefined. Setting the derivative of the given function equal to zero, we get 11eb651f_caea_8253_b0a1_69f5fee97a3b_TB8255_11, which gives two critical points, 11eb651f_caea_824e_b0a1_7d34ea4f10e4_TB8255_11 and 11eb651f_caea_824f_b0a1_9547bf655353_TB8255_11. The second derivative is 11eb651f_caea_8254_b0a1_4a2edaaf8ab8_TB8255_11. 
At 11eb651f_caea_824e_b0a1_7d34ea4f10e4_TB8255_11, the second derivative is positive, indicating a relative minimum.
At 11eb651f_caea_824f_b0a1_9547bf655353_TB8255_11, the second derivative is negative, indicating a relative maximum.
Therefore, the answer is (B).

Question 5

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; relative maximum value: 
B)  ; relative minimum value: 
C)  ; saddle point: 
D) there are no critical points

Accepted Answer

The answer of Find the critical point(s) of the function....

Question 6

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; relative maximum value: 
B)  and  ; relative maximum value:  ; relative minimum value: 
C)  ; saddle point: 
D)  and  ; saddle point:  ; relative minimum value: 
E) there are no critical points

Accepted Answer

The answer of Find the critical point(s) of the function....

Question 7

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If the data consist of two distinct points, then the least-squares line is just the line that passes through the two points.

Accepted Answer

The answer of Determine whether the statement is true or...

Question 8

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  is the point of maximum, 11 is the relative maximum
B)  is the critical point, it is impossible to determine the relative extrema of the function
C)  is the point of minimum, -11 is the relative minimum
D)  is the critical point, the function has neither a relative maximum nor a relative minimum at this point
E) no critical points

Accepted Answer

To find the critical point(s), we need to find where the derivative of the function equals zero or is undefined. Taking the derivative of the function, we get:

f'(x) = -3x^2 + 22x - 21

Setting f'(x) equal to zero and solving for x, we get:

-3x^2 + 22x - 21 = 0

Using the quadratic formula, we get:

x = (22 ± sqrt(22^2 - 4(-3)(-21)))/(-6) = (11 ± sqrt(31))/3

Therefore, the critical point is x = (11 ± sqrt(31))/3.

To classify the nature of this critical point, we need to use the second derivative test. Taking the second derivative of the function, we get:

f''(x) = -6x + 22

Plugging in the critical point x = (11 ± sqrt(31))/3, we get:

f''((11 + sqrt(31))/3) = 22 - 6(11 + sqrt(31))/3 = -2 - 2sqrt(31)
f''((11 - sqrt(31))/3) = 22 - 6(11 - sqrt(31))/3 = -2 + 2sqrt(31)

Since f''((11 + sqrt(31))/3) and f''((11 - sqrt(31))/3) are both negative, the function is concave down at both critical points. Therefore, (11 ± sqrt(31))/3 are both points of relative maximum and minimum, respectively.

Plugging in x = (11 + sqrt(31))/3 into the original function, we get:

f((11 + sqrt(31))/3) = -1/3(11 + sqrt(31))^3 + 11/2(11 + sqrt(31))^2 - 21(11 + sqrt(31))/3 + 22 = -34.593

Plugging in x = (11 - sqrt(31))/3 into the original function, we get:

f((11 - sqrt(31))/3) = -1/3(11 - sqrt(31))^3 + 11/2(11 - sqrt(31))^2 - 21(11 - sqrt(31))/3 + 22 = -37.407

Therefore, the relative minimum is f((11 + sqrt(31))/3) = -34.593 and the relative maximum is f((11 - sqrt(31))/3) = -37.407. The best choice is C.

Question 9

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; relative minimum value: 
B)  ; relative maximum value: 
C)  ; saddle point: 
D) there are no critical points

Accepted Answer

The given function has a critical point at x = -2. To classify the nature of this point, we need to use the second derivative test:

f'(x) = 3x^2 + 6x - 9
f''(x) = 6x +6

At x = -2, we have f'(-2) = 3(-2)^2 + 6(-2) - 9 = -9 < 0 and f''(-2) = 6(-2) + 6 = -6 < 0. This means that the critical point is a saddle point.

Therefore, the answer is (C). The function has a saddle point at x = -2.

Question 10

Find the equation of the least-squares line for the given data. Draw a scatter diagram for the given data and graph the least-squares line. ​  ​
A) ​ 
B) ​ 
C) ​ 
D) ​

Accepted Answer

The answer of Find the equation of the least-squares line...

Question 11

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. The least-squares line must pass through at least one data point. A) It is false. Example. Find the least-squares line for the data Solution. Here, we have n=5 and The least-squares line for the data is given by linear equation y= f(x)= mx+b where the constants m and b satisfy the normal equations Then, we obtain the normal equations Solving them, we found m=1, b=0.4 Therefore, the required least-squares line is y= x+0.4 The scatter diagram and the least-squares line are shown in the figure. We can see that the line does not pass through any data point. B) It is true, Suppose that we are given two data points If we try to fit a straight line to these data points, the line will miss the first and the second data points by the amounts d₁, d₂, respectively. The principle of least squares states that the straight line L that fits the data points best is the one chosen by requiring that the sum of the squares be made as small as possible. But it is possible only in case when least-squares line passes through at least one data point.

Accepted Answer

As shown in the example, it is possible for the least-squares line to not pass through any of the data points. Therefore, the statement is false.

Question 12

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; relative minimum value: 
B)  ; relative maximum value: 
C)  ; saddle point:

Accepted Answer

To find critical points, we need to find where the first derivative equals zero or is undefined. In this case, the first derivative is:
f'(x) = 3x^2 - 4x
Setting f'(x) = 0, we get:
3x^2 - 4x = 0
x(3x - 4) = 0
So the critical points are x = 0 and x = 4/3.

To classify the nature of each point using the second derivative test, we need to find the second derivative:
f''(x) = 6x - 4
Plugging in our critical points, we get:
f''(0) = -4 < 0, so the critical point at x = 0 is a local maximum.
f''(4/3) = 2 > 0, so the critical point at x = 4/3 is a local minimum.

Therefore, the best choice is C, which correctly identifies the critical point at x = 4/3 as a saddle point, which is incorrect.

Question 13

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ,  are the critical points, the function has neither a relative maximum nor a relative minimum
B)  is the critical point, the function has neither a relative maximum nor a relative minimum
C)  is the critical point, the function has neither a relative maximum nor a relative minimum
D)  ,  are the critical points, the function has neither a relative maximum nor a relative minimum
E) no critical points

Accepted Answer

The answer of Find the critical point(s) of the function....

Question 14

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; relative maximum value: 
B)  ; relative maximum value: 
C)  ; relative maximum value:

Accepted Answer

To find the critical point(s), we set the derivative equal to 0 and solve for x:
f'(x) = 3x^2 - 8x = x(3x-8) = 0
x = 0 or x = 8/3 are the critical points.
To classify the nature of each critical point, we need to take the second derivative:
f''(x) = 6x - 8
At x = 0, f''(0) = -8, which is negative, so the critical point at x = 0 is a relative maximum.
At x = 8/3, f''(8/3) = 2, which is positive, so the critical point at x = 8/3 is a relative minimum.
Therefore, the best choice is B, which correctly identifies the critical point at x = 0 as a relative maximum. Choices A and C incorrectly identify the critical points at x = 0 and x = 8/3, respectively.

Question 15

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; saddle point: 
B)  ; relative maximum value: 
C)  ; relative minimum value:

Accepted Answer

The critical point 11eb651f_cae9_e5fb_b0a1_05f928c9f964_TB8255_11 is the only point given and we can use the second derivative test to classify its nature. Taking the second derivative of the provided function, we get f''(x) = 2 - 12x. Plugging in the critical point, we get f''(11eb651f_cae9_e5fb_b0a1_05f928c9f964_TB8255_11) = -20 which is negative. This means that the critical point is a relative minimum value. So the answer is only C.

Question 16

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; relative minimum value: 
B)  ; relative maximum value: 
C)  ; saddle point: 
D) there are no critical points

Accepted Answer

The answer of Find the critical point(s) of the function....

Question 17

Find an equation of the least-squares line for the data. Please round the coefficients to the nearest hundredth. x
1
1
2
3
4
4
5
Y
2
3
5
6
6
6)5
8)5
​
A) 
B) ​ 
C) ​ 
D) ​ 
E) ​

Accepted Answer

The answer of Find an equation of the least-squares line...

Question 18

Find the equation of the least-squares line for the given data. Draw a scatter diagram for the given data and graph the least-squares line. ​  ​
Please round the coefficients in your equation to three decimal places.
​
A) ​ 
B) ​ 
C) ​ 
D) ​

Accepted Answer

The answer of Find the equation of the least-squares line...

Question 19

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; relative minimum value: 
B)  ; saddle point: 
C)  ; relative maximum value: 
D) there are no critical points

Accepted Answer

The answer of Find the critical point(s) of the function....

Question 20

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A)  ; saddle point: 
B)  ; relative maximum value: 
C)  ; relative minimum value: 
D) there are no critical points

Accepted Answer

The critical point can be found by setting the derivative equal to zero:

f'(x) = 3x^2 - 6x = 0

Solving for x, we get x = 0 or x = 2.

To classify the nature of each critical point, we need to take the second derivative:

f''(x) = 6x - 6

Plugging in x = 0, we get f''(0) = -6, which means the point x = 0 is a relative maximum value.

Plugging in x = 2, we get f''(2) = 6, which means the point x = 2 is a relative minimum value.

Therefore, the answer is C.

Quiz 8: Calculus of Several Variables

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If the data consist of two distinct points, then the least-squares line is just the line that passes through the two points.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the equation of the least-squares line for the given data. Draw a scatter diagram for the given data and graph the least-squares line.

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. The least-squares line must pass through at least one data point.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find an equation of the least-squares line for the data. Please round the coefficients to the nearest hundredth. x 1 1 2 3 4 4 5 Y 2 3 5 6 6 6) 5 8) 5

Find the equation of the least-squares line for the given data. Draw a scatter diagram for the given data and graph the least-squares line. Please round the coefficients in your equation to three decimal places.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Quiz 8: Calculus of Several Variables

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If the data consist of two distinct points, then the least-squares line is just the line that passes through the two points.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the equation of the least-squares line for the given data. Draw a scatter diagram for the given data and graph the least-squares line. ​ ​

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. ​ The least-squares line must pass through at least one data point.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find an equation of the least-squares line for the data. Please round the coefficients to the nearest hundredth. x 1 1 2 3 4 4 5 Y 2 3 5 6 6 6) 5 8) 5 ​

Find the equation of the least-squares line for the given data. Draw a scatter diagram for the given data and graph the least-squares line. ​ ​ Please round the coefficients in your equation to three decimal places. ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​ ​

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the equation of the least-squares line for the given data. Draw a scatter diagram for the given data and graph the least-squares line.

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. The least-squares line must pass through at least one data point.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find an equation of the least-squares line for the data. Please round the coefficients to the nearest hundredth. x 1 1 2 3 4 4 5 Y 2 3 5 6 6 6) 5 8) 5

Find the equation of the least-squares line for the given data. Draw a scatter diagram for the given data and graph the least-squares line. Please round the coefficients in your equation to three decimal places.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.