💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

# Suppose $f"$ is continuous on $(-\infty, \infty)$.(a) If $f'(2) = 0$ and $f"(2) = -5$, what can you say about $f$?(b) If $f'(6) = 0$ and $f"(6) = 0$, what can you say about $f$?

## (a) By the Second Derivative Test, if $f^{\prime}(2)=0$ and $f^{\prime \prime}(2)=-5<0, f$ has a local maximum at $x=2$(b) If $f^{\prime}(6)=0,$ we know that $f$ has a horizontal tangent at $x=6 .$ Knowing that $f^{\prime \prime}(6)=0$ does not provide any additionalinformation since the Second Derivative Test fails. For example, the first and second derivatives of $y=(x-6)^{4}$$y=-(x-6)^{4},$ and $y=(x-6)^{3}$ all equal zero for $x=6,$ but the first has a local minimum at $x=6,$ the second has alocal maximum at $x=6,$ and the third has an inflection point at $x=6$

Derivatives

Differentiation

Volume

### Discussion

You must be signed in to discuss.
##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp

### Video Transcript

this funk are in this problem that are second derivative F double prime of X is continuous for all real numbers or from negative infinity to infinity. And in part they were told that F prime of two is equal to zero, n. f double prime of two is equal to negative five. And were asked, what can we say about this function given this information? Well, since we know that F prime of two is equal to zero, that means that we have a critical point At X is equal to two, which means that X is equal to two. We could have a local maximum or minimum value. And since we know the sign of our second derivative at this point X is equal to two, we can tell whether this is a maximum or minimum. Since this sign of our second derivative at X is equal to two is negative. So since second derivative at X equal to two is less than zero. That means that this is a local maximum. So we know that we have a local maximum and this is from the second derivative test At X is equal to two. If the second derivative was greater than zero, then it would have been a local minimum. So what we can tell from F prime of two being equal to zero and F double prime of two being equal to negative five is that we have a local maximum at X is equal to two. And now for part B we're told that F prime of X or F prime of six is equal to zero and F double prime of X. Sorry, for double prime of six Is also equal to zero. So we know that there is a critical point, Since our derivative is equal to zero At this point X is equal to six. So critical point, uh X is equal to six. And since our second derivative at X or at six is equal to zero, um That means that this isn't a local maximum or a local minimum for it to be a local maximum minimum, this double prime of six would have had to been positive or negative. But since it's equal to zero, that means at this point At f double prime of X or f double prime of six could be an inflection point. So potential inflection point at X is equal to six, but it is not a local maximum minimum, since our second derivative isn't positive or negative.

Oregon State University

#### Topics

Derivatives

Differentiation

Volume

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

Lectures

Join Bootcamp