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nignag [31]
2 years ago
13

A function is given. (a) Find all the local maximum and minimum values of the function and the value of x at which each occurs.

(b) Find the intervals on which the function is increasing and on which the function is decreasing. State each answer correct to two decimal places. U (x) = 4 (x3 - x)

Mathematics
1 answer:
Allisa [31]2 years ago
8 0

Answer:

a)

x = -1/√3 = -0.58    is a local maximum

x = 1/√3  =  0.58     is a local minimum

b)

(-1/√3 , 1/√3 ) DECREASING

(-∞, -1/√3) U (1/√3, ∞)    INCREASING

Step-by-step explanation:

To answer all of the questions we must obtain the derivative of the fucntion:

If

U(x) = 4(x^3 - x)

then

U'(x) = 4(3x^2 - 1)

U''(x) = 4(3*2 x) = 24 x

U'''(x) = 24

The local maxima and minima of the function U(x) can be found when U'(x) = 0

this occurs when :

3x^2 = 1

that is:

x = ±1/√3

We will know if they are a minimum or a maximum evaluating this points on the second derivative (you can look for it as <u><em>Second derivative test</em></u>), if the result is positive the point corresponds to a minimum and if it is negative it will be a maximum.

Here it is easy to determine wheather is a maximum or a minimum because the second derivative is 24x, therefore if x is negative or positive so the second derivative will be.

a)

x = -1/√3 = -0.58    is a local maximum

x = 1/√3  =  0.58     is a local minimum

b)

the intervals at which the function is increasing { decreasing } is given when the first derivative is positive { negative }

The first derivative will be positive when:

3x^2 > 1

|x| > 1/√3  -->  x > 1/√3   and    x < -1/√3

The first derivatice will be negative when:

3x^2 < 1

|x| < 1/√3  -->  x < 1/√3   and    x > -1/√3

Therefore the intervals are:

(-1/√3 , 1/√3 ) DECREASING

(-∞, -1/√3) U (1/√3, ∞)    INCREASING

<u><em>** The attached image is a plot of the fucntion where you can see part of the intervals and the local maximum and minimum</em></u>

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Answer:

T = 40 (minutes?)

Step-by-step explanation:

Given the equation G = 50 + 20T, where G = the total amount of gas and T = the amount of Time, we know that there is already 50 gallons of gas in the tank and that the rate at which it is pumped is 20 gallons per minute (I am assuming, though it is not stated in the problem).  Given that the tank holds 850 gallons, we can plug in the values into the equation and solve for the missing variable, T:

850 = 50 + 2T

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3 years ago
What is the value can u show work
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3 years ago
The midpoint of a circle is (3,2), the
Marysya12 [62]

yo sup??

mid point of circle=centre of circle

let centre be O(3,2)

and the point be P(-4,8) and the other point be Q(x,y)

by mid point formula

-4+x/2=3 and 8+y/2=2

x=10 and y=-4

therefore coordinates of the other point are (10,-4)

Hope this helps.

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3 years ago
................. plz let only deltaalex answer:) don’t trust anyone on here but them!
nevsk [136]

Answer:

Step-by-step explanation:

oh law of sines   soo  sin(A) / a = Sin(B) / b

2a)

here DE = side f   then let's find that side with law of sines and plug side f in as 'a' in law of sines , so

Sin(28) / a = Sin(54) / 31  ( where a = f = DE )

can you do the algebra here Emily?  or do you want me to show that?

Sin(28) *31 / Sin(54) = a

17.989 =a

18.0 =a       (rounded to nearest 10th)

DE = 18.0 m

2b)

find RT  where RT = side a

since they give use 2 of the other sides of the triangle we can quickly figure out the S angle.   180 = 71+62+S   so  S = 47°  ( I'm going kinda fast b/c I don't know how much time you have  :? )

then Sin(A) /a = Sin(B) / b

Sin(47) / a = Sin(62) / 29

Sin(47) *29 / Sin(62) = a

24.020 = a

RT = 24.0 meters  ( rounded to nearest 10th )

2c)

they want us to find ∠X with law of sines soo

I'll just use the letter they give us for variables

Sin(X) / x  = Sin(Y ) / y

Sin(X) = x* Sin(y) / y

X = arcSin [ 14 * Sin(108) / 34 ]

X = arcSin [ 0.3916115]

X = 23.0548°

X= 23°  ( rounded to nearest degree )

2d)

they ask us to find ∠E

I'll , again, use the letter they give us for variables

Sin(E) / e = Sin( F ) / f

Sin(E) / 18 = Sin(78) / 30

E = arcSin[ 18 * Sin(78) / 30 ]

E = arcSin [ 0.5868856 ]

E = 35.936 °

E = 36 °    ( rounded to nearest degree )

:)

3 0
3 years ago
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