Intervals concave up and concave down: g(x)=(3x-2)^5

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jenna89

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Good Afternoon,
Directions:
Find the intervals on which the following functions are concave up and concave down.
problem: g(x)=(3x-2)^5

My work
g(x)= 5(3-2)4 (3)
15(3x-2)^4=0
15=0 no solution
3x-2=0
x=2/3

Therefore, the critical point is x=2/3

Concaves up since x is higher than 0

Please help!
 
Re: Intervals concave up and concave down

jenna89 said:
Find the intervals on which the following function [is] concave up and concave down.

g(x)=(3x-2)^5

My work

g'(x)= 5(3x - 2)^4 (3) (very sloppy typing or lack of knowledge)

15(3x-2)^4=0

15=0 no solution Huh?

3x-2=0
x=2/3

Therefore, the critical point is x=2/3 (This result makes sense to me, but it only earns you little partial credit.)

Concaves up since x is higher than 0 This is your answer to the given exercise?
Click to expand...


Read the given exercise!

It requires an INTERVAL over which g(x) is concave up. Where is your interval?

It requires an INTERVAL over which g(x) is concave down. Where is your statement about concavity downward?

Start by reviewing the definition of concavity. Look up the test for concavity; it requires first determining the second derivative.

Use the critical point to determine the intervals to test.

Please reply if you need help understanding the definition of concavity and/or how to implement the second-derivative test for concavity.

Cheers,

~ Mark :)

 
Re: Intervals concave up and concave down

Thanks Mark..This is what I came up with
Find the intervals on which the following function [is] concave up and concave down.

g(x)=(3x-2)^5
g'(x)= 5(3x - 2)^4 (3)
15(3x-2)^4=0
15=0 no solution Huh?
3x-2=0 x=2/3

2nd derivative
g''(x) 60 (3x-2)^3(3)
180(3x-2)^3=0

g(2/3)=(3x-2)^5
(3*2/3-2)^5
(2-2)=0

Inflection point is (2/3,0)
greater than 2/3 concave up
lower than 2/3 concave down

Thanks in advance for your help
 
Intervals and concavity: g(x)=(3x-2)^5

Find the intervals on which the following function [is] concave up and concave down.

g(x)=(3x-2)^5
g'(x)= 5(3x - 2)^4 (3)
15(3x-2)^4=0
15=0 no solution Huh?
3x-2=0 x=2/3

2nd derivative
g''(x) 60 (3x-2)^3(3)
180(3x-2)^3=0

g(2/3)=(3x-2)^5
(3*2/3-2)^5
(2-2)=0

Inflection point is (2/3,0)
greater than 2/3 concave up
lower than 2/3 concave down

Thanks in advance for your help
 
Re: Intervals and concavity

jenna89 said:
Find the intervals on which the following function [is] concave up and concave down.

g(x)=(3x-2)^5
g'(x)= 5(3x - 2)^4 (3)
15(3x-2)^4=0
15=0 no solution Huh?
3x-2=0 x=2/3

2nd derivative
g''(x) 60 (3x-2)^3(3)
180(3x-2)^3=0

g(2/3)=(3x-2)^5
(3*2/3-2)^5
(2-2)=0

Inflection point is (2/3,0)
(What is )greater than 2/3 concave up
(What is )lower than 2/3 concave down

I assume you are supposed write it mathematical notations like:

f(x) concave up for x > 2/3

Thanks in advance for your help
Click to expand...

duplicate post:

viewtopic.php?f=3&t=31322&p=120908#p120908
 
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