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nignag [31]
3 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]3 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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yaroslaw [1]

Answer:

New mean length of these 17 people=11.62

Step-by-step explanation:

<em>Step 1: Determine the total length of children's fingers</em>

L=l×n

where;

L=total length (cm)

l=mean length (cm)

n=number of children

replacing;

L=6.7×6=40.2 cm

The total children's fingers length=40.2 cm

<em>Step 2: Determine the total length of adults fingers</em>

total length of adults fingers=mean length of adults fingers×number of adults

where;

mean length of adults=14.3

number of adults=11

replacing;

total length of adults fingers=(14.3×11)=157.3 cm

<em>Step 3: Determine the new mean length</em>

new mean length=total length of adults and children fingers/number of adults and children

where;

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replacing;

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Answer:

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Step-by-step explanation:

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To find:

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Solution:

<em>Domain of a function </em>is defined as the set of valid input values that can be given to the function for which the function is defined.

Here, input is time.

We can not have negative values for time.

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Maximum value for time can be 2.236 seconds.

Therefore the domain is:

<em>Interval notation: [0, 2.236]</em>

<em>Set Builder notation: </em>\{t\ |\ 0\le t\le 2.236\}

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B.

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