The graph of f(x)= 3/1+x^2 is shown in the figure to the right. Use the second derivative of f to find the intervals on which f is concave upward or concave downward and to find the inflection points of f.

The Graph Of F(x)= 3/1+x^2 Is Shown In The Figure To The Right. Use The Second Derivative Of F To Find

Answers

Answer 1

Answer:

Concave Up Interval: [tex](- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)[/tex]

Concave Down Interval: [tex](\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )[/tex]

General Formulas and Concepts:

Calculus

Derivative of a Constant is 0.

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Quotient Rule: [tex]\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]

Chain Rule: [tex]\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Second Derivative Test:

Possible Points of Inflection (P.P.I) - Tells us the possible x-values where the graph f(x) may change concavity. Occurs when f"(x) = 0 or undefinedPoints of Inflection (P.I) - Actual x-values when the graph f(x) changes concavityNumber Line Test - Helps us determine whether a P.P.I is a P.I

Step-by-step explanation:

Step 1: Define

[tex]f(x)=\frac{3}{1+x^2}[/tex]

Step 2: Find 2nd Derivative

1st Derivative [Quotient/Chain/Basic]:                           [tex]f'(x)=\frac{0(1+x^2)-2x \cdot 3}{(1+x^2)^2}[/tex]Simplify 1st Derivative:                                                           [tex]f'(x)=\frac{-6x}{(1+x^2)^2}[/tex]2nd Derivative [Quotient/Chain/Basic]:     [tex]f"(x)=\frac{-6(1+x^2)^2-2(1+x^2) \cdot 2x \cdot -6x}{((1+x^2)^2)^2}[/tex]Simplify 2nd Derivative:                                                       [tex]f"(x)=\frac{6(3x^2-1)}{(1+x^2)^3}[/tex]

Step 3: Find P.P.I

Set f"(x) equal to zero:                    [tex]0=\frac{6(3x^2-1)}{(1+x^2)^3}[/tex]

Case 1: f" is 0

Solve Numerator:                           [tex]0=6(3x^2-1)[/tex]Divide 6:                                          [tex]0=3x^2-1[/tex]Add 1:                                              [tex]1=3x^2[/tex]Divide 3:                                         [tex]\frac{1}{3} =x^2[/tex]Square root:                                   [tex]\pm \sqrt{\frac{1}{3}} =x[/tex]Simplify:                                          [tex]\pm \frac{\sqrt{3}}{3} =x[/tex]Rewrite:                                          [tex]x= \pm \frac{\sqrt{3}}{3}[/tex]

Case 2: f" is undefined

Solve Denominator:                    [tex]0=(1+x^2)^3[/tex]Cube root:                                   [tex]0=1+x^2[/tex]Subtract 1:                                    [tex]-1=x^2[/tex]

We don't go into imaginary numbers when dealing with the 2nd Derivative Test, so our P.P.I is [tex]x= \pm \frac{\sqrt{3}}{3}[/tex] (x ≈ ±0.57735).

Step 4: Number Line Test

See Attachment.

We plug in the test points into the 2nd Derivative and see if the P.P.I is a P.I.

x = -1

Substitute:                    [tex]f"(x)=\frac{6(3(-1)^2-1)}{(1+(-1)^2)^3}[/tex]Exponents:                   [tex]f"(x)=\frac{6(3(1)-1)}{(1+1)^3}[/tex]Multiply:                        [tex]f"(x)=\frac{6(3-1)}{(1+1)^3}[/tex]Subtract/Add:              [tex]f"(x)=\frac{6(2)}{(2)^3}[/tex]Exponents:                  [tex]f"(x)=\frac{6(2)}{8}[/tex]Multiply:                       [tex]f"(x)=\frac{12}{8}[/tex]Simplify:                       [tex]f"(x)=\frac{3}{2}[/tex]

This means that the graph f(x) is concave up before [tex]x=\frac{-\sqrt{3}}{3}[/tex].

x = 0

Substitute:                    [tex]f"(x)=\frac{6(3(0)^2-1)}{(1+(0)^2)^3}[/tex]Exponents:                   [tex]f"(x)=\frac{6(3(0)-1)}{(1+0)^3}[/tex]Multiply:                       [tex]f"(x)=\frac{6(0-1)}{(1+0)^3}[/tex]Subtract/Add:              [tex]f"(x)=\frac{6(-1)}{(1)^3}[/tex]Exponents:                  [tex]f"(x)=\frac{6(-1)}{1}[/tex]Multiply:                       [tex]f"(x)=\frac{-6}{1}[/tex]Divide:                         [tex]f"(x)=-6[/tex]

This means that the graph f(x) is concave down between  and .

x = 1

Substitute:                    [tex]f"(x)=\frac{6(3(1)^2-1)}{(1+(1)^2)^3}[/tex]Exponents:                   [tex]f"(x)=\frac{6(3(1)-1)}{(1+1)^3}[/tex]Multiply:                       [tex]f"(x)=\frac{6(3-1)}{(1+1)^3}[/tex]Subtract/Add:              [tex]f"(x)=\frac{6(2)}{(2)^3}[/tex]Exponents:                  [tex]f"(x)=\frac{6(2)}{8}[/tex]Multiply:                       [tex]f"(x)=\frac{12}{8}[/tex]Simplify:                       [tex]f"(x)=\frac{3}{2}[/tex]

This means that the graph f(x) is concave up after [tex]x=\frac{\sqrt{3}}{3}[/tex].

Step 5: Identify

Since f"(x) changes concavity from positive to negative at [tex]x=\frac{-\sqrt{3}}{3}[/tex] and changes from negative to positive at [tex]x=\frac{\sqrt{3}}{3}[/tex], then we know that the P.P.I's [tex]x= \pm \frac{\sqrt{3}}{3}[/tex] are actually P.I's.

Let's find what actual point on f(x) when the concavity changes.

[tex]x=\frac{-\sqrt{3}}{3}[/tex]

Substitute in P.I into f(x):                    [tex]f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+(\frac{-\sqrt{3} }{3} )^2}[/tex]Evaluate Exponents:                          [tex]f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }[/tex]Add:                                                    [tex]f(\frac{-\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }[/tex]Divide:                                                [tex]f(\frac{-\sqrt{3}}{3} )=\frac{9}{4}[/tex]

[tex]x=\frac{\sqrt{3}}{3}[/tex]

Substitute in P.I into f(x):                    [tex]f(\frac{\sqrt{3}}{3} )=\frac{3}{1+(\frac{\sqrt{3} }{3} )^2}[/tex]Evaluate Exponents:                          [tex]f(\frac{\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }[/tex]Add:                                                    [tex]f(\frac{\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }[/tex]Divide:                                                [tex]f(\frac{\sqrt{3}}{3} )=\frac{9}{4}[/tex]

Step 6: Define Intervals

We know that before f(x) reaches [tex]x=\frac{-\sqrt{3}}{3}[/tex], the graph is concave up. We used the 2nd Derivative Test to confirm this.

We know that after f(x) passes [tex]x=\frac{\sqrt{3}}{3}[/tex], the graph is concave up. We used the 2nd Derivative Test to confirm this.

Concave Up Interval: [tex](- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)[/tex]

We know that after f(x) passes [tex]x=\frac{-\sqrt{3}}{3}[/tex] , the graph is concave up until [tex]x=\frac{\sqrt{3}}{3}[/tex]. We used the 2nd Derivative Test to confirm this.

Concave Down Interval: [tex](\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )[/tex]

The Graph Of F(x)= 3/1+x^2 Is Shown In The Figure To The Right. Use The Second Derivative Of F To Find

Related Questions

Please answer the screenshot I attached below

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Answer:

i think the answer is

[tex] \frac{1}{ {x}^{ \frac{4}{3} }} [/tex]

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Answer:

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Answers

Answer:

-2x + 2

Step-by-step explanation:

2x - 3 - 2y - 4x + 2y + 5

2x - 4x + 2y - 2y + 5 - 3

-2x + 2

29-3x=5x+5


Can you work it out

Answers

Answer:

x=3

Step-by-step explanation:

29 - 3x = 5x + 5

Subtract 5 from both sides:

24 - 3x = 5x

Add 3x to both sides:

24 = 8x

Divide both sides by 8:

X = 3


At a local baseball game, the concession stand has two different meal choices. You can get 7 hotdogs and 5
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per hotdog:

per drink:

Answers

Answer:

Each hotdog is $3 and each drink is $4.

Step-by-step explanation:

Let h be the number of hotdogs and d the number of drinks,

7h + 5d = 41

4h + 8d = 44

Let's eliminate h.  Multiply the first by 4 and the second by 7.

28h + 20d = 164

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Answers

Answer:

4

Step-by-step explanation:

4 x 5= 20

4x6 = 24

So 4 can go in 20 and 24 making it the common factor

The common factor of 20 and 24 would be; 4

How to find Greatest common factor?

Greatest common factor, as the nathe me implies, is greatest among all common factors among the quantities given. We can factorize the quantities in smallest relative factors possible (one level of factors over which all quantities can form themselves), and collect as many common factors as we could. Those will together greatest GCF(Greatest common factor).

We need to find the common factors of 20 and 24

Therefore,

Factors are;

4 x 5= 20

4 x 6 = 24

Hence, we can conclude that 4 can go in 20 and 24. it is the common factor.

The common factor of 20 and 24 would be; 4

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Answers

Answer: 3.65

Step-by-step explanation:

This question is asking basically what is 50% or 1/2 of 7.3.

Let's expand the number 7.3

Then, we end up getting 7.30.

We have to find 1/2 of 7.30.

[tex]\frac{1}{2}[/tex] of 7.30 = [tex]3.65[/tex]

Hope this helps you!

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Answer:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

$0.40

Step-by-step explanation:

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On a coordinate plane, the x-axis is labeled Number of Days and the y-axis is labeled Number of Computers. A line goes through points (3, 36), (4, 48), (5, 60). The graph shows a proportional relationship between the number of computers produced at a factory per day. In three days, 36 computers are produced; 48 computers are produced in 4 days; and 60 computers are produced in 5 days. Find the unit rate of computers per day using the graph. Unit rate: computers per day

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Step-by-step explanation:

Answer:

12

Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Answers

Answer:

20r + 310

Step-by-step explanation

For the first part.

The required expression is 20x + 310 which represents the area of the park as the sum of the areas of the sports field and the playground.

What is the expression?

Expressions are defined as mathematical statements that have a minimum of two terms containing variables or numbers.

Let x represent the length in meters of the sports field.

Here, the area of the sports field = 20 × x = 20x units

And, the area of the playground = 15.5 × 20 = 310 units

Now, the required expression to represent the area of the park will be :

⇒ area of the sports field + area of the playground

⇒ 20x + 310

Thus, the required expression is 20x + 310.

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Use an exponent to write each expression 3/4 x 3/4 x 3/4 x 3/4 x 3/4

Answers

Answer:

[tex] { \frac{3}{4} }^{5} [/tex]

The given expression 3/4 x 3/4 x 3/4 x 3/4 x 3/4 can be written as (3/4)⁵ in the form of an exponent.

What is the expression?

Expressions are defined as mathematical statements that have a minimum of two terms containing variables or numbers.

An exponent is a shorthand way of representing a repeated multiplication of the same number.

As per the given question, the number being multiplied is 3/4, and it is being multiplied by itself 5 times. The exponent, 5, tells us how many times the base, 3/4, is being multiplied by itself.

⇒ 3/4 x 3/4 x 3/4 x 3/4 x 3/4

⇒ (3/4)⁵

Therefore, the given expression can be written as (3/4)⁵.

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Steph bought a used bicycle restored it and then marked it up by 1/4 of the price she paid later she sold it for the marked-up price of $80 how much in dollar did steph pay for the bicycle ?

Answers

Answer: would be $64

Answer:

Steph payed $64 for the bicycle

Step-by-step explanation:

80/5=16

16×4=64

hopefully this helps :)

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Answers

Answer:

-45

Step-by-step explanation:

9x5=45, and since there is a negative, you put it before 45.

Hope I helped!

Please mark Brainliest!!!

A theater has 384 seats. Each row has 16 seats. The area model represents this situation.What is the value of x in the area model? Enter your answer in the box.​

Answers

Answer:

X = 4 because 20 times 16 is 320 than if you add 4 to the 20 you will get 24. 24 times 16 is 384. BOom thats your answer and i know because i just answered it.

Step-by-step explanation:

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-40 4 10 20 I think this is the answer I’m not sure

7(3x -4) = 49



can you help me

Answers

Answer:

3 2/3

Step-by-step explanation:

7(3x-4)=49

21x-28=49

21x=77

x=3 2/3

Solve each of the following:
a. 3 movie tickets cost $36. At this rate, what is the cost per ticket?
b. 3 ice cream cones cost $8.25. At this rate, how
much do 2 ice cream cones cost?
C-3 bananas cost $0.99. At this rate, how much do 5 bananas cost?
a. $
b. $

Answers

Step-by-step explanation:

A. 36÷3=12 $12 per ticket

B. 8.25÷3=2.75 $2.75 per ice-cream cone so do 2.75×2=5.50 $5.50 for 2 ice creams

C. 0.99÷3=0.33 $0.33 per banana so do 0.33×5=1.65

$1.65 for 5 bananas

HOPE THIS HELPS

Jan tossed a coin 50 times and got a head 20 times. What is the
experimental probability? What is the theoretical probability?

Answers

Answer:

The standard answer for this is 50%.  

Step-by-step explanation:

  However, this is based on the implicit assumption that the coin is fair.

 

If there are reasonable grounds to doubt that the coin is fair, the theoretical probability must be based on observed statistics.   In this scenario, the probability is 60%.  

I need help please Will mark brainliest. Thank you so much.

Answers

The answer is, 168.....(extra characters)

4x - 4 - 8y + 8 - 5x + 1 =

Answers

Answer:

-x - 8y + 5

Step-by-step explanation:

Add the numbers

4x - 4 - 8y + 8 - 5x + 1 =

4x + 5 - 8y - 5y

Combine like terms

4x + 5 - 8y - 5y

-x + 5 - 8y

Rearrange terms

-x + 5 - 8y

-x - 8y + 5

The solution of expression is,

⇒ - x - 8y + 5

We have to given that,

An expression is,

⇒ 4x - 4 - 8y + 8 - 5x + 1

Now, We can simplify by combining like terms as,

⇒ 4x - 4 - 8y + 8 - 5x + 1

⇒ 4x - 5x - 8y - 4 + 8 + 1

⇒ - x - 8y + 5

Therefore, The solution of expression is,

⇒ - x - 8y + 5

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Help me with this please it says how can you isolate the term f/3 in f/3 + 22 = 17

Answers

Answer:

Given:

[tex]\sf {\dfrac {f}{3}}+22=17 [/tex]

To find:-

[tex]\sf {\dfrac {f}{3}}[/tex]

Solution:-First write the equation

[tex]{:}\dashrightarrow[/tex][tex]\sf {\dfrac {f}{3}}+22=17 [/tex]

Now take 22 to the other side by which the "+"sign will turn "-".

[tex]{:}\dashrightarrow[/tex][tex]\sf {\dfrac {f}{3}}=17-22 [/tex]

Simplify

[tex]{:}\dashrightarrow[/tex][tex]\sf {\dfrac {f}{3}}=-5 [/tex]

[tex]\therefore[/tex][tex]{\underline{\boxed{\bf {\dfrac {f}{3}}=(-5)}}}[/tex]

Answer: Subtract 22 from both sides.

Step-by-step explanation:

f/3 +22 = 17

f/3 +22 -22 = 17 -22  Subtracting the same value from both sides keeps the equation equal.  

Then simplify. The +22 and -22= 0 They "cancel"   17-22 = -5

f/3 = -17  f/3 is now isolated-- by itself-- in the equation.

If you had to solve for f, the next step would be to multiply both sides by  3.

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