Average rate of change. Need help setting up the equation PLEASE!
What is the average rate of change of f(x), represented by the graph, over the interval [-1, 2]
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OpenStudy (kitkat16):
OpenStudy (ageta):
little help @sleepyjess
OpenStudy (sleepyjess):
do you know what "over the interval [-1, 2]" means?
OpenStudy (kitkat16):
my understandingis reading the graph to set up the equation.
OpenStudy (kitkat16):
f(b)-f(a)/b-a
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OpenStudy (kitkat16):
Im pretty sure I deive each term but this graph is confusing me
OpenStudy (kitkat16):
*divide
OpenStudy (sleepyjess):
are you supposed to use the x values of -1 and 2?
OpenStudy (kitkat16):
yes x/y
OpenStudy (sleepyjess):
@ganeshie8
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OpenStudy (kitkat16):
\[f(2)-f(-1)/2-(-1)\]
OpenStudy (kitkat16):
Is that how you set it up?
OpenStudy (sleepyjess):
is there an equation?
OpenStudy (kitkat16):
no thats another problem. :/
OpenStudy (anonymous):
You don't actually need it since you can use the information provided by the graph. However, the function depicted is: \(\displaystyle f(x) = \frac{5}{x}\)
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OpenStudy (kitkat16):
(2.5-5)/(2-(-5) My problem is understand how to read the graph
OpenStudy (kitkat16):
wouldn't it be -5?
if so then x=-5/2.5
then xwould = -2
OpenStudy (kitkat16):
@surjithayer can u help me please?
OpenStudy (kitkat16):
you line -1 up with -5 yes? and then line up 2 with 2.5 yes? Not sure if I am doing this correctly.
OpenStudy (kitkat16):
@NoelGreco are u busy?
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OpenStudy (kitkat16):
2.5-(-5)/2-(-1) = 2.5
OpenStudy (kitkat16):
@Loser66 can you help me please?
OpenStudy (kitkat16):
omg I hate math!!!!
OpenStudy (kitkat16):
now Im even more confused. how do I set up this problem do you know?
OpenStudy (kitkat16):
2.5-5/((2-(-5)/0-2)) :(
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OpenStudy (kitkat16):
yes
OpenStudy (loser66):
That's why I said I hate this problem.
OpenStudy (kitkat16):
so could that be cancled out
OpenStudy (kitkat16):
I don't know probably not
OpenStudy (kitkat16):
so was I totally wrong on the steps to do this problem?
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OpenStudy (kitkat16):
im watching a video on it now. Thanks for trying to help.