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Show that if $f(x) = x^4$, then $f"(0) = 0$, but $(0, 0)$ is not an inflection point of the graph of $f$.

No infection point because $f \prime \prime(x)$ is always positive

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Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Michael J.

Idaho State University

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Video Transcript

we're being the Astra showed that if every Mexican, except for and then we're told that then f double prints were reserved and security was not, in fact, a point. The craft back. So why is that the case? So let's go ahead and addresses. So the second driven of this function Medical two. Twelve x squared. And so if you plug in zero, you get dio. And if you recall, you may think that the definition is oven infected points where the second derivative equal to zero and that's not. And that's entirely wrong. But it's also not entirely correct, because one of the other definition of an inflection point is there has to be a change of sign occurring around F, occurring around the inflection points around change of fine around inflection point. And so if you look at this function, it is twelve x squared and twelve X squared means that this baby dysfunction right here is always increasing. It is always always greater than zero is always positive. So this con caked up everywhere and so is not changing con cavity. And there's no point in which changes time cavity because there's always Khan came up so I mean, there is no inflection points, and that's why. And that's why it is important to check the sign and just look at the function. Sometimes. Look for squares, too. I mean squares and powers of even number because generally those air bit more tricky. And so there is no infection point because there is no sign change, no signed change, that's it.

Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Michael J.

Idaho State University

Lectures

Join Bootcamp