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Mathematics 18 Online
OpenStudy (canada907cat):

The graph of y = f (x) is shown below on the domain [0, 5]. Sketch the graph of y = f ′′(x) on the same domain.

OpenStudy (canada907cat):

OpenStudy (canada907cat):

@zepdrix

satellite73 (satellite73):

probably a line

satellite73 (satellite73):

this looks like a cubic the first derivative would be a quadratic the second derivative would be a line

OpenStudy (canada907cat):

So what would that look like? @satellite73

zepdrix (zepdrix):

Upload a picture or something :P I don't want to download your dirty virus filled pdf file XD lol

OpenStudy (canada907cat):

Lol okay let me attempt to do that.

OpenStudy (canada907cat):

OpenStudy (canada907cat):

I am not sure how to make it look bigger.

satellite73 (satellite73):

cubic with positive leading coefficient first derivative is a parabola where \(f\) is increasing \(f'\) is positive, then where \(f\) is deceasing, \(f'\) is negative, then \(f'\) is positive again draw a nice parabola for that one

satellite73 (satellite73):

btw is it clear than "positive" is a synonym for "above the x axis"?

OpenStudy (canada907cat):

Okay so would it start from 3 and up from there? Sorry Im am not good at explaining this.

satellite73 (satellite73):

the zeros of your parabola will be the local max and min of your function, namely 1 and 3

zepdrix (zepdrix):

|dw:1475194331176:dw|We have an inflection point somewhere around here, the graph changes from CCD to CCU. So our second derivative will be zero at this location.

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