Question

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

Answer 1

y = x^5-80x+100

Answer 2

i think it has to do with finding the f''(x)=0

Answer 3

I know that

Answer 4

y'(x) = 5x^4 -80

Answer 5

\[y' = 5x^4 -8\]
\[y'' = 20x^3\]

Answer 6

y''(x) = 20x^3

Answer 7

set y'' = 0 and solve.

Answer 8

i got 0... so i wasn't sure if that's right

Answer 9

0 = 20x^3

Answer 10

Yes.

Answer 11

Okay thanks!

Answer 12

Oh i forgot, is it the local minimum (x=0)?

Answer 13

@abb0t

Answer 14

If the second derivative is zero then the critical point can be anything. Look @ the graph of ur function.

Answer 15

so it's just the local extrema is located at x=0

Answer 16

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?

Answer 17

Only if it's zero.
if < 0 = relative min
if > 0 = relative max.

Answer 18

ohhhh ok! Thank you so much!