Question

Please tell me if i have this right. Find all critical numbers and use second derivative test to determine all local extrema

Answer 1

\[f(x)=\frac{ x ^{2}-1 }{ x }\]

Answer 2

Here we go again :)

Answer 3

For the derivative i got \[f'(x)=\frac{ 1 }{ x }+1\]

Answer 4

sorry x should be squared on the derivative

Answer 5

So it is undefined when x=0, so this is the critical number?

Answer 6

therefore with the second derivative being \[-\frac{ 2 }{ ^{x ^{3}} }\]

Answer 7

it would also be undefined, so the second derivative test fails in this case?

Answer 8

I simplified the original equation to this:
\[f(x) = x - \frac{ 1 }{ x }\]
so:
\[f'(x) = 1 + \frac{ 1 }{ x ^{2} }\]
When you set the first derivative test equal to zero, you get no solutions, so there are NO critical points.
Second derivative test is used to find points of inflection, not critical points/extrema. So, in the function there are no critical points. x = 0 is undefined.

Answer 9

no, sorry the first equation is \[\frac{ 1 }{ x ^{2} }+1\]

Answer 10

but i guess follows same thing as you discussed

Answer 11

I thought that an undefined point was also a critical point?

Answer 12

It depends on how you define a critical point in your class. Lots of people do. In which clase, x = 0 is a critical point because it is non-differentiable.

Answer 13

So you are saying the answer could be said that the first derivative test fails therefore second can't even be done, or critical point is 0 but second derivative test fails?

Answer 14

No. The first derivative test fails, so there are no extrema. There could still be points of inflection, so it's necessary to try the second derivative test. In this case, the second derivative test fails as well, so there are no points of inflection and no extrema.

Answer 15

ok gotcha, looking at the graph i can see that

Answer 16

Awesome! :)