Find the average rate of change of the function from x1 to x2

[Heinzelmannchen]

f(x)= -3x^2 - x from x1=5 to x2=6 Use inclement .1 as the rate of change. Make a table.

The inclement and making a table part of the problem are confusing me.


i figured i need to evaluate for f(5), and f(6) then find the slope.

f(5)=-80 f(6)=-114 -114- (-80)/ 6-5= -34


Any advice for what i'm missing? I don't know what it means by make a table and use inclement of .1

 

[Deleted member 4993]

f(x)= -3x^2 - x from x1=5 to x2=6 Use inclement .1 as the rate of change. Make a table.

The inclement and making a table part of the problem are confusing me.


i figured i need to evaluate for f(5), and f(6) then find the slope.

f(5)=-80 f(6)=-114 -114- (-80)/ 6-5= -34


Any advice for what i'm missing? I don't know what it means by make a table and use inclement of .1

I assume the problem meant increment and not inclement.

If it is increment then, you need to find values of f(5.0), f(5.1), f(5.3)......f(5.7), f(5.8), f(5.9) & f(6.0)

 

[lookagain]

f(x)= -3x^2 - x from x1=5 to x2=6 Use > > > inclement < < < .1 as the rate of change. Make a table.

The > > > inclement < < < and making a table part of the problem are confusing me.


i figured i need to evaluate for f(5), and f(6) then find the slope.

> > > f(5)=-80 f(6)=-114 -114- (-80)/ 6-5= -34 < < <


Any advice for what i'm missing? I don't know what it means by make a table and use inclement of .1

I don't know whether that word is supposed to be "increment" or "inclement."

That aside, *space out* your work horizontally and vertically, and use needed grouping symbols

around your numerator and denominator.



For example, you could have typed the highlighted portion from the quote box as:


f(5) = -80\(\displaystyle \ \ \ \ \) f(6) = -114


[-114 - (-80)]/(6 - 5) =

(-114 + 80)/(1) =

-34

 

[Heinzelmannchen]

I don't know whether that word is supposed to be "increment" or "inclement."

That aside, *space out* your work horizontally and vertically, and use needed grouping symbols

around your numerator and denominator.



For example, you could have typed the highlighted portion from the quote box as:


f(5) = -80\(\displaystyle \ \ \ \ \) f(6) = -114


[-114 - (-80)]/(6 - 5) =

(-114 + 80)/(1) =

-34

You're right, I realize how sloppy it looks now. I was rushing because i had work soon, i'll make sure to look it over before posting next time. Any input if i tell you the word is inclement?

I assume the problem meant increment and not inclement.

If it is increment then, you need to find values of f(5.0), f(5.1), f(5.3)......f(5.7), f(5.8), f(5.9) & f(6.0)

No the word is INCLEMENT. My teacher is from England so that might be part of it. I'm assuming it has to do with slope.

 

[Heinzelmannchen]

NO, it's NOT INCLEMENT, it's incRement.
Why don't you take the time to look it up
If that's from your teacher, then the teacher made a typo.

What makes you assume it has to do with slope? Your problem is not slope related....

The word Inclement threw me off, it made me assume it has to do with slope. You have to admit, if you're stupid like me and don't know what the word means it sounds slope related. You're right though, stupid of me to overlook the teacher making a mistake. It's just he is very strict so i won't be able to talk to him about it until he grades it. :| So did i do the problem then m=-34? I just need to make a table for the values of f(5.1), f(5.2)...f(6) right ? The average should still be the same


I appreciate the help thus far guys. Please excuse my ignorance .

 

[Deleted member 4993]

The word Inclement threw me off, it made me assume it has to do with slope. You have to admit, if you're stupid like me and don't know what the word means it sounds slope related. You're right though, stupid of me to overlook the teacher making a mistake. It's just he is very strict so i won't be able to talk to him about it until he grades it. :| So did i do the problem then m=-34? I just need to make a table for the values of f(5.1), f(5.2)...f(6) right ? The average should still be the same


I appreciate the help thus far guys. Please excuse my ignorance .

The average will not be the same.

 

[stapel]

The word Inclement threw me off, it made me assume it has to do with slope.

"IncLement" (inn-KLEMM-int) is generally used in relation to weather. For instance, if it's "raining cats and dogs", then the weather is said to be "incLement".

"IncRement" (INN-kruh-ment) is generally used in relation to change, usually of the consistent sort. For instance, if it's been "raining cats and dogs" for hours, perhaps days, one might refer to the water climbing toward one's house as approaching in one-meter-a-day increments.

 

[Heinzelmannchen]

The average will not be the same.

Ok, assuming i set it up right then i just find the values of f(5), f(5.1), f(5.2)...f(6) right? I then find the sum of them all and divide by 11. I'm very grateful for Ti-89's


f(5)+f(5.1)+f(5.2)...+f(6)= -1062.46

Average rate of change = ( -1062.46) / 11 = -96.5873

I think i got it now. Thanks again for the help guys

"IncLement" (inn-KLEMM-int) is generally used in relation to weather. For instance, if it's "raining cats and dogs", then the weather is said to be "incLement".

"IncRement" (INN-kruh-ment) is generally used in relation to change, usually of the consistent sort. For instance, if it's been "raining cats and dogs" for hours, perhaps days, one might refer to the water climbing toward one's house as approaching in one-meter-a-day increments.

I did actually look the word up after posting this, thanks for clearing that up though. Good analogy by the way

 

[Deleted member 4993]

Ok, assuming i set it up right then i just find the values of f(5), f(5.1), f(5.2)...f(6) right? I then find the sum of them all and divide by 11. I'm very grateful for Ti-89's


f(5)+f(5.1)+f(5.2)...+f(6)= -1062.46

Average rate of change = ( -1062.46) / 11 = -96.5873 ........................ Incorrect

What you have calculated is average f(x) - not rate of change of f(x)

I think i got it now. Thanks again for the help guys


I did actually look the word up after posting this, thanks for clearing that up though. Good analogy by the way

If you need to find average rate of change - then you need to find [{f(5.1)-f(5)}/0.1], [{f(5.2)-f(5.1)}/0.1], [{f(5.3)-f(5.2)}/0.1],....[{f(6)-f(5.9)}/0.1] and take average of that.

 

[Heinzelmannchen]

Geezzzz Heinzel, go stand in the corner for 27 minutes...

Thanks that inspires me to keep doing math problems. I apologize for exhibiting human characteristics and making a mistake.

If you need to find average rate of change - then you need to find [{f(5.1)-f(5)}/0.1], [{f(5.2)-f(5.1)}/0.1], [{f(5.3)-f(5.2)}/0.1],....[{f(6)-f(5.9)}/0.1] and take average of that.

I obviously don't expect anyone to find the average, but does -31 sound about right? After adding all the averages of delta y i get -31.



For the table i have [f(x+h)-f(x)]/h | ΔY

 

[Heinzelmannchen]

Don't get a heart attack, Heinzel: that was a joke...

I figured it was but wasn't sure. I aint even mad

http://en.wikipedia.org/wiki/Poe's_law