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Mathematics 19 Online
OpenStudy (anonymous):

How do I find the critical numbers of f' from the graph of f''?

OpenStudy (anonymous):

OpenStudy (anonymous):

I know how to find the critical numbers for the graph of f from f', but this is asking for the critical numbers of f' from f"", which I don't know how to do.

OpenStudy (anonymous):

^*f"

OpenStudy (anonymous):

if you know how to find the critical numbers of f with information from f', then you can apply that same knowledge with the second derivative and apply it to f'... put it this way... let g=f' then g' = f'' if you know the where g' is zero, that's the critical numbers of g...

OpenStudy (anonymous):

RIght, I know that where ever x=0 on the graph of f', then those values are critical numbers of f. But how do I go about finding the critical numbers of f' just by looking at the graph of f''?

OpenStudy (anonymous):

let f' = g, so f'' = g' ... g' = 0 means f'' = 0. so "whereever g'=0 that's the critical numbers for g." same statement: "whereever f''=0 that's the critical numbers for f'."

OpenStudy (anonymous):

So it's kind of the same process as finding the critical numbers of f from the graph of f''? The critical numbers of f' are where x=0 on f''?

OpenStudy (anonymous):

*from the graph of f'

OpenStudy (anonymous):

^type

OpenStudy (anonymous):

*typo

OpenStudy (anonymous):

exactly...

OpenStudy (anonymous):

So since x=0 four times on the graph of f'', then there are four critical numbers of f'?

OpenStudy (anonymous):

yes...

OpenStudy (anonymous):

Ah. I guess I was just expecting the process to be different for second derivatives and such. Well thanks.

OpenStudy (anonymous):

btw... you mean f''=0 four times on the graph, then there are four critical numbers of f'

OpenStudy (anonymous):

Oh yes, that's what I mean. Thanks.

OpenStudy (anonymous):

yw...)

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