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Problem 87 Medium Difficulty

Prove that if $ (c, f(c)) $ is a point of inflection of the graph of $ f $ and $ f" $ exists in an open interval that contains $ c $, then $ f"(c) = 0 $. [Hint: Apply the First Derivative Test and Fermat's Theorem to the function $ g = f' $.]

Answer

By hypothesis $g=f^{\prime}$ is differentiable on an open interval containing $c$. since $(c, f(c))$ is a point of inflection, the concavity
changes at $x=c,$ so $f^{\prime \prime}(x)$ changes signs at $x=c .$ Hence, by the First Derivative Test, $f^{\prime}$ has a local extremum at $x=c$
Thus, by Fermat's Theorem $f^{\prime \prime}(c)=0$

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Video Transcript

So let's suppose see, FFC is an inflection point of our graph of our function of F. Of X. And so what we're gonna do is since we know that our point CFC is a inflection point, we can use the fact that the con cavity is going to change at X equals C. So this means than khan cavity changes at X is equal to see. And wherever the con cavity changes by the first derivative test, so can say by the first derivative test, our first derivative F prime of C will be a local extreme. Er that means it will be either a local max or minimum value for our um for our first derivative of death. And since our first derivative F has a local extreme A. At this point, C. We can say that by Fermat's theorem, which just says that a local minimum or maximum of a function F means that the derivative of that function is zero at that point. So if this F prime of C has a local extreme A. At that point, see that means the derivative of F prime, which should be F double prime C is equal to zero, so the derivative of F prime of C Is equal to zero. And this is just the second derivative of F is equal to zero at sea. So now we have proved that if CFC is an inflection point of fx, then that means that the con cavity changes at that point, which means that our first derivative is going to have a local extreme of their macs are men, which by Fairmont's thera means that our second derivative is going to be equal to zero.