Question

The curve y = ax^3 - 24x + b has a locale minimum at (2;-17)

9.1) Calculate the values of a and b
9.2) The co-ordinate of the other turning point , E , on the curve
9.3) For what value of k wil ax^3 - 24x + b = k have three different roots.
9.4) Determine the co-ordinate of the inflection point of the curve

Answer 1

Do you have any idea about how to proceed?

Answer 2

I can do everything except 9.3.....

Answer 3

Great! What have you found for a and b?

Answer 4

My book just gives the answer k is an element of (-17;47)

Answer 5

The answer to a and b appearently is a = 2 and b = 15

Answer 6

Well, if we differentiate \(y = ax^3-24x+b\) we get \[\frac{dy}{dx} = 3ax^2-24\] and differentiating again get \(6ax\). Setting the first derivative equal to 0 and plugging in our known value of x,
\[0 = 3a(2)^2-24\]\[24 = 12a\]\[a=2\]and we can find \(b\) by using the original equation and (2,-17).

But you knew that... I'm a little confused by part c, I must admit, because it looks to me like there are many values of k that give 3 different roots...

Answer 7

Oh, I didn't see that you had posted the answer. Yes, those two values make you get only two distinct roots. But how to find them...hmm...

Answer 8

I dont understand it at all, I did "mistype" though it says, for what "values" not value...

Answer 9

Hahaha... I should note that if you calculate the maximum and the minumum... you get the x y value of the minumum and the y value of the maximum to be respectively -17 and 47..... I still dont see the relation though....

Answer 10

Ah, duh, of course! At those spots you have an inflection point, and that means a repeated root.

Answer 11

I was looking at a much more difficult approach :-)

If we solve \[2x^3-24x+15 == k\] the roots would be:

\[\left\{\left\{x\to \frac{4\ 2^{2/3}}{\left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}+\frac{\left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}{2^{2/3}}\right\},\left\{x\to -\frac{2\ 2^{2/3} \left(1+i \sqrt{3}\right)}{\left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}-\frac{\left(1-i \sqrt{3}\right) \left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}{2\ 2^{2/3}}\right\},\left\{x\to -\frac{2\ 2^{2/3} \left(1-i \sqrt{3}\right)}{\left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}-\frac{\left(1+i \sqrt{3}\right) \left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}{2\ 2^{2/3}}\right\}\right\}\] (plus two more that aren't coming over) and trying to eyeball how we would make two of the three be equal was giving me quite a headache!

Answer 12

Root 1: \[\frac{4\ 2^{2/3}}{\left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}+\frac{\left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}{2^{2/3}}\]

Root 2: \[-\frac{2\ 2^{2/3} \left(1+i \sqrt{3}\right)}{\left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}-\frac{\left(1-i \sqrt{3}\right) \left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}{2\ 2^{2/3}}\]

Root 3: \[-\frac{2\ 2^{2/3} \left(1-i \sqrt{3}\right)}{\left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}-\frac{\left(1+i \sqrt{3}\right) \left(-15+k+\sqrt{-799-30 k+k^2}\right)^{1/3}}{2\ 2^{2/3}}\]

Answer 13

wow.... Is that a formula?..

Answer 14

If we set k = 47, they evaluate to 4, -2, and -2.

If we set k = -17, they evaluate to 2, -4, and 2.

Anyone can see that ;-)

Answer 15

So, yeah, the local minima or maxima was the key. When a polynomial has an even number of roots at the same point, it is either a local minima or maxima. If it has an odd number, then it is an inflection point.

Answer 16

Here are some graphs. No interaction with me is complete without some graph :-)

You can see that one graph (k = -17) has the curve dipping down and touching the x-axis at x = 2, so that's our repeated root. The other graph (k = 47) has the curve rising up and touching the x-axis at x = -2 for our other repeated root.

Answer 17

And for completeness, here's another graph where we don't have a duplicate root, and you see the curve crossing the x-axis 3 times: