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OpenStudy (anonymous):
what is the formula of f' and f''? the graph is posted below.
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OpenStudy (anonymous):
OpenStudy (amistre64):
just follow the rules of differentation and derive them :)
what part is giving you troubles?
myininaya (myininaya):
so f is given right?
Where do you see that the graph has horizontal tangents?
myininaya (myininaya):
remember when you graph f' you are graphing the slope for each point
i always start by seeing where my horizontal tangents are (where the slope is 0)
OpenStudy (amistre64):
i think its gonna be a bit more convoluted than that if you try to do it graphwise
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myininaya (myininaya):
if the slope is 0, then f'=0 there
if the slope is positive, then f'>0 there
if the slope is negative, then f'<0 there
OpenStudy (amistre64):
[t e^t]' = t e^t + e^t
[-e^(t^2)]' = -2t e^(t^2) ; this one id have to wonder about
[+3]' = 0
myininaya (myininaya):
oh oops they give the equation lol
myininaya (myininaya):
or i mean the function
OpenStudy (amistre64):
lol ... its readable as long as you aint on an iphone :)
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myininaya (myininaya):
gj amistre
OpenStudy (amistre64):
if i see it right:
\[f'(t) = t \ e^t + e^t -2t\ e^{t^2} \]
OpenStudy (anonymous):
e^t (t+1) - 2e^t^2 t
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