Finding a for local minimum on y = -4x

[Bushy]

For $$ f(x) = (x-1)^2(x-a) $$

Find the exact value of $$a$$ such that the local min lies on the point with equation $$y=-4x
$$

 

[Greg]

Hi bushy. Are you sure you have the question typed correctly? Points have coordinates, lines have equations. That aside, what have you tried?

 

[Bushy]

Well the question comes with a diagram. The linear equation goes through that minimum.

What have I tried?

I have made the cubic equal to the linear equation but that gives me a value for x in terms of a

 

[Bushy]

I found it $$a = \frac{1}{2}(5+3\sqrt{5})$$

 

[MarkFL]

I think what I would, is first compute the first derivative:

$$y'=(x-1)(3x-(2a+1))$$

And then the second derivative:

$$y''=6x-2(a+2)$$

Okay, now let's look at the root of the first derivative:

$$x=\frac{2a+1}{3}$$

With this value for $x$, we require the second derivative to be positive:

$$6\left(\frac{2a+1}{3}\right)-2(a+2)>0$$

$$2a+1-a-2>0$$

$$a>1$$

This tells us that if the local minimum is going to occur anywhere other than that $(1,0)$, then we need $1<a$. And when we have $1<a$, the local minimum is locate at:

$$(x,y)=\left(\frac{2a+1}{3},\frac{4}{27}(1-a)^3\right)$$

Now, in order for this point to lie on the line $y=-4x$, we require:

$$\frac{4}{27}(1-a)^3=-4\left(\frac{2a+1}{3}\right)$$

$$(a-1)^3=9(2a+1)$$

$$(a+2)\left(a^2-5a-5\right)=0$$

Taking the only root such that $1<a$, we obtain:

$$a=\frac{5+3\sqrt{5}}{2}$$

And this agrees with the value you found. :D