Problem

For what values of $ a $ and $ b $ is $ (2, 2.5) …

05:25

Problem 75 Hard Difficulty

(a) If the function $ f(x) = x^3 + ax^2 + bx $ has the local minimum value $ - \frac{2}{9} \sqrt{3} $ at $ x = 1/ \sqrt{3} $, what are the values of $ a $ and $ b $?
(b) Which of the tangent lines to the curve in part (a) has the smallest slope?

Answer

(a) $f(x)=x^{3}+a x^{2}+b x \Rightarrow f^{\prime}(x)=3 x^{2}+2 a x+b . \quad f$ has the local minimum value $-\frac{2}{9} \sqrt{3}$ at $x=1 / \sqrt{3},$ so
\[
f^{\prime}\left(\frac{1}{\sqrt{3}}\right)=0 \Rightarrow 1+\frac{2}{\sqrt{3}} a+b=0 \quad \text { (1) and } \quad f\left(\frac{1}{\sqrt{3}}\right)=-\frac{2}{9} \sqrt{3} \Rightarrow \frac{1}{9} \sqrt{3}+\frac{1}{3} a+\frac{1}{3} \sqrt{3} b=-\frac{2}{9} \sqrt{3}
\]
Rewrite the system of equations as
\[
\begin{aligned}
\frac{2}{3} \sqrt{3} a+& b=-1 \\
\frac{1}{3} a+\frac{1}{3} \sqrt{3} b &=-\frac{1}{3} \sqrt{3}
\end{aligned}
\]
and then multiplying (4) by $-2 \sqrt{3}$ gives us the system
\[
\begin{aligned}
\frac{2}{3} \sqrt{3} a+b &=-1 \\
-\frac{2}{3} \sqrt{3} a-2 b &=2
\end{aligned}
\]
Adding the equations gives us $-b=1 \Rightarrow b=-1 .$ Substituting -1 for $b$ into (3) gives us
\[
\frac{2}{3} \sqrt{3} a-1=-1 \Rightarrow \frac{2}{3} \sqrt{3} a=0 \Rightarrow a=0 . \text { Thus, } f(x)=x^{3}-x
\]
(b) To find the smallest slope, we want to find the minimum of the slope function, $f^{\prime},$ so we'll find the critical
\[
\text { numbers of } f^{\prime} . \quad f(x)=x^{3}-x \Rightarrow f^{\prime}(x)=3 x^{2}-1 \Rightarrow f^{\prime \prime}(x)=6 x . \quad f^{\prime \prime}(x)=0 \Leftrightarrow x=0
\]
At $x=0, y=0, f^{\prime}(x)=-1,$ and $f^{\prime \prime}$ changes from negative to positive. Thus, we have a minimum for $f^{\prime}$ and
$y-0=-1(x-0),$ or $y=-x,$ is the tangent line that has the smallest slope.

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Video Transcript

But if the function has been making excuses, expert has local minimum value at negative to three other nine at X equals one of every three one of the values of A and B. And then we're asked to find the tangent line in the current wish. Which point? And it has the smallest. So So they're a couple of information that are key Teo solving this. So we know that of one over three is equal to negative, too. Three night over nine. And so we can ask you plug this into our equation right away. So we get one over three cube, Of course, A terms. One over three. Uh, this could be squared. And in class B b times one of every three. And it is all equal to negative, too. Uh, three over nine. And then once we Cuba, we get won over three Route three and then plus a over three. And then we have B over route three. And then this is still equal to negative too, with three over nine. So now what we're going to do to simplify this more we're gonna move by both sides by Route three. Don't close side five three. And once we do that, we get one plus a with three plus tree be and this is equal to negative too. So this is our first equation. So what we're going to be doing instead of using the systems of question? So how we're gonna get a next equation or we take the derivative, It could tell us information about the derivative. We know that there is some sort of local minimum value occurring and show in local miniatures. We know that prime of X is equal to zero. So well, first, what we have to get to do it is a three x Square plus three B. And I know about that three experience bliss to a eggs A plus B. And we know that at crime of one over three or that's equal to zero. So we're gonna plug that in zero equals three times one over three squared. Tow us to a tryingto won over three cluster B, and that's all. You go to zero. Won't you simplify this. You get one glass to a three. Presby equals zero. Now we're going to do is multiplied by three on both side and then this will give us another equation. Three a bluff too, eh? Three. Trust me, baby is equal to zero. And then you could say about this further to a three plus three. B is equal to negative three. Um, now we're going to do this is our second equation. Now, over. When they do, it's going to attract the first equation from our second equation. And so what? That gives us a result. It gives us that a route three Deco to zero and then we know that a because zero. And now that we have a we gonna see plug this into any of our equations over this case, we'LL do this one. This will give us Ah, I'm writing it in Red Hook A zero plus three V you could negatively now could solve for B we'Ll be just negative one. And that is our answer for a for the second part, Um, we have to just take the Now that we know the values, we could just take the derivative So primal backs Well, we know that on a zero. So this doesn't exist and then be physical too negative one. And that's just going to go away in a derivative. So it's just gonna come out to be six ex, actually. And so where, which is Attendant Line has a smaller stop with the small stuff is where Empathy for the zero. And so people are in antiquity because you're well, that occurred. That exit call Tio. So that is the point in which our mama's little girl

Fahad P.

Related Courses

Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 3

How Derivatives Affect the Shape of a Graph

Related Topics

Derivatives

Differentiation

Volume

Top Calculus 2 / BC Educators

Heather Z.

Oregon State University

Kristen K.

University of Michigan - Ann Arbor

Samuel H.

University of Nottingham

Joseph L.

Boston College