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Illusion [34]
3 years ago
9

Consider the function f ( x ) = − 2 x 3 + 27 x 2 − 108 x + 9 . For this function there are three important open intervals: ( − [

infinity] , A ) , ( A , B ) , and ( B , [infinity] ) where A and B are the critical numbers. Find A and B
Mathematics
1 answer:
Darya [45]3 years ago
3 0

Answer:

<em>A=3 and B=6</em>

Step-by-step explanation:

<u>Increasing and Decreasing Intervals of Functions</u>

Given f(x) as a real function and f'(x) its first derivative.

If f'(a)>0 the function is increasing in x=a

If f'(a)<0 the function is decreasing in x=a

If f'(a)=0 the function has a critical point in x=a

As we can see, the critical points may define open intervals where the function has different behaviors.

We have

f ( x ) = - 2 x^3 + 27 x^2 - 108 x + 9

Computing the first derivative:

f' ( x ) = - 6 x^3 + 54 x - 108

We find the critical points equating f'(x) to zero

- 6 x^3 + 54 x - 108=0

Simplifying by -6

x^2 -9 x +18=0

We get the critical points

x=3,\ x=6

They define the following intervals

(-\infty,3),\ (3,6),\ (6,+\infty)

Thus A=3 and B=6

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Find the vertex, focus, directrix, and focal width of the parabola. (5 points)
Anarel [89]

\bf \textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf -\cfrac{1}{20}x^2=y\implies \cfrac{x^2}{-20}=y\implies x^2=-20y\implies (x-0)^2=-20(y-0) \\\\\\ (x-\stackrel{h}{0})^2=4(\stackrel{p}{-5})(y-\stackrel{k}{0})

something noteworthy is that the squared variable is the "x", thus the parabola is a vertical one, the "p" value is negative, so is opening downwards, and the h,k is pretty much the origin,

vertex is at (0,0)

the focus point is "p" or 5 units down from there, namely at (0, -5)

the directrix is "p" units on the opposite direction, up, namely at y = 5

the focal width, well, |4p| is pretty much the focal width, in this  case, is simply yeap, you guessed it, 20.

8 0
3 years ago
8 - 4y + (-2y) + 5 <br><br><br> What is this answer please tell me
Irina-Kira [14]

Answer:

<em>13 - 6y </em>

Step-by-step explanation:

8 - 4y + (-2y) + 5 = <em>13 - 6y</em>

8 0
3 years ago
Read 2 more answers
Please help!
UkoKoshka [18]

The values of x in the triangles and the angles in the rhombus are illustrations of tangent ratios

  • The values of x in the triangles are 21.4 units, 58 degrees and 66 degrees
  • The angles in the rhombus are 44 and 46 degrees, respectively

<h3>How to determine the values of x?</h3>

<u>Triangle 1</u>

The value of x is calculated using the following tangent ratio

tan(25) = 10/x

Make x the subject

x = 10/tan(25)

Evaluate

x = 21.4

<u>Triangle 2</u>

The value of x is calculated using the following tangent ratio

tan(x) = 8/5

Evaluate the quotient

tan(x) = 1.6

Take the arc tan of both sides

x = arctan(1.6)

Evaluate

x = 58

<u>Triangle 3</u>

The value of x is calculated using the following tangent ratio

tan(x) = 0.34/0.15

Evaluate the quotient

tan(x) = 2.27

Take the arc tan of both sides

x = arctan(2.27)

Evaluate

x = 66

<h3>How to calculate the angles of the rhombus?</h3>

The lengths of the diagonals are:

L1 = 2 in

L2 = 5 in

Represent the angles with x and y.

The measures of the angles are calculated using the following tangent ratios

tan(0.5x) = 2/5 and y = 90 - x

Evaluate the quotient

tan(0.5x) = 0.4

Take the arc tan of both sides

0.5x = arctan(0.4)

Evaluate

0.5x = 22

Divide by 0.5

x = 44

Recall that:

y = 90 - x

This gives

y = 90 - 44

Evaluate

y = 46

Hence, the angles in the rhombus are 44 and 46 degrees, respectively

Read more about tangent ratio at:

brainly.com/question/13347349

5 0
2 years ago
What is the value of x
likoan [24]

Answer: The value of x is 8x

Step-by-step explanation:

3 0
4 years ago
How do I solve and graph this?
lianna [129]
3x -y ⩾ 6

3x - 6 ⩾ y

now, with inequalities, what we do is, we graph the line of 3x - 6 = y, and then we shade the "true region".

if we pick a point on say hmmm (4, 0), namely x = 4 and y = 0, we can plug that in the inequality and see what we get,

3(4) - 0 ⩾ 6

12 - 0 ⩾ 6

12 ⩾ 6   

is 12 really greater or equals to 6?  well yes, therefore, the point (4, 0) lies on the "true region", since it's true, 12 is indeed ⩾ 6, so, where that point is, we shade.

now, the ⩾ means equals to or greater, and therefore, since the values could also equal the boundary points, the line is a solid line, because it includes the line itself, as well as the shading.

check the picture below.

3 0
3 years ago
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