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Second derivative test to find the local extrema for the function?

Mathematics
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y = x^5-80x+100
i think it has to do with finding the f''(x)=0
I know that

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Other answers:

y'(x) = 5x^4 -80
\[y' = 5x^4 -8\] \[y'' = 20x^3\]
y''(x) = 20x^3
set y'' = 0 and solve.
i got 0... so i wasn't sure if that's right
0 = 20x^3
Yes.
Okay thanks!
Oh i forgot, is it the local minimum (x=0)?
If the second derivative is zero then the critical point can be anything. Look @ the graph of ur function.
so it's just the local extrema is located at x=0
and if it's like... 3 and -3 as the local extrema. I don't know if it's a local max or min correct?
Only if it's zero. if < 0 = relative min if > 0 = relative max.
ohhhh ok! Thank you so much!

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