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Max and Min of Functions without Derivative

Date: 04/22/2003 at 11:36:29
From: Roy
Subject: Max and Min of functions without derivative

Hello Dr. Math,

I was curious to know if there is a general way to find the max and 
min of cubic functions without using derivatives. Our book does this 
with the use of graphing calculators, but I was wondering if there is 
a way to find the critical points without derivatives. Any help is 
greatly appreciated!

Thanks,
Roy Delgado 


Date: 04/22/2003 at 12:36:43
From: Doctor Peterson
Subject: Re: Max and Min of functions without derivative

Hi, Roy.

Yes, there is a way, and it may be very instructive!

Look at the graph of a cubic, and recall that if a polynomial has a 
double root, it will be tangent to the x-axis (at q here):

                        |               *
                        |
                        |
                        |
                 *      |              *
             *        * |
                        |
           *            *            *
                        |
                        | *        *
    ------*-------------+------*---------------
          p             |      q
                        |
                        |
                        |
                        |
         *              |
                        |
                        |
                        |
                        |

So given a general cubic, if we shift it vertically by the right 
amount, it will have a double root at one of the turning points.

So, given an equation

  y = ax^3 + bx^2 + cx + d

any turning point will be a double root of the equation

  ax^3 + bx^2 + cx + d - D = 0

for some D, meaning that that equation can be factored as

  a(x-p)(x-q)^2 = 0

Equating these and expanding, we have

                          a(x-p)(x-q)^2 = ax^3 + bx^2 + cx + d - D

                a(x-p)(x^2 - 2qx + q^2) = ax^3 + bx^2 + cx + d - D

   a(x^3 - (p+2q)x^2 + q(2p+q)x - pq^2) = ax^3 + bx^2 + cx + d - D

  ax^3 - a(p+2q)x^2 + aq(2p+q)x - apq^2 = ax^3 + bx^2 + cx + d - D

Since these must be equal for ALL x, we can equate coefficients:

         a = a
  -a(p+2q) = b
  aq(2p+q) = c
    -apq^2 = d-D

We have three equations (ignoring the first, which is simply the 
reason I multiplied my factored form by a in the first place) in three 
unknowns, p, q, and D, so we can solve to find the double root q, 
which is the location of the turning point.

To do this, we'll eliminate p by solving the second equation above 
for p:

  p = -(b/a + 2q)

and putting this into the third equation:

  aq(-2(b/a + 2q) + q) = c

This simplifies to

     -2bq - 3aq^2 = c

  3aq^2 + 2bq + c = 0

(Note that this is the derivative of the cubic we are working with. 
The rest of the work is just what we would do if we were using 
calculus, but with different reasoning.)

Now we solve this for q using the quadratic formula:

      -2b +- sqrt(4b^2 - 12ac)   -b +- sqrt(b^2 - 3ac)
  q = ------------------------ = ---------------------
               6a                         3a

This gives both turning points, since there are two ways to make a 
double root. Our last equation gives the value of D, the y-coordinate 
of the turning point:

  D = apq^2 + d = -a(b/a + 2q)q^2 + d = -2aq^3 - bq^2 + d

    = (aq^3 + bq^2 + cq + d) - (3aq^2 + 2bq + c)q

    = aq^3 + bq^2 + cq + d

(since 3aq^2 + 2bq + c = 0), as we would expect given that x = q;
so we don't really have to carry out this step.

This is the sort of work that had to be done before calculus was 
invented. In many cases, calculus is really just a shortcut for 
algebra.

Thanks for asking the question, because I'd never considered 
approaching it without calculus before! If you have any further 
questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 04/22/2003 at 12:47:08
From: Roy
Subject: Thank you (Max and Min of functions without derivative)

Thanks for the quick response, Dr. Peterson. That is just what I was 
looking for! I think my students will definitely benefit from seeing 
this approach and appreciate it once they get into calculus.

Roy Delgado

Date: 06/25/2016 at 10:19:17
From: Yousuf
Subject: Maximum and minimum of a cubic function WITHOUT calculus

Hi Doctor Math,

In the conversation above, you said this:

   Any turning point will be a double root of the equation

      ax^3 + bx^2 + cx + d - D = 0

   for some D, meaning that that equation can be factored as

      a(x - p)(x - q)^2 = 0

The '- D' is a little confusing to understand as to where you got it from.
How (or where?) did you obtain it to form that first equation? What
significance does it play in this analogy?

I think I do understand the basic gist of what you're trying to achieve
here. I understand the analogy that (I THINK) we picture first a general
cubic graph which has no turning points touching the x-axis. If we can
translate such a graph by the right amount vertically, what we get is a
cubic graph where one of the turning points is touching the x-axis, making
the axis a tangent to that point (we get a double root).

So with this said (daft as it may seem, but I may have answered my own
question), is the 'D' the difference of the vertical component of the
vector used to translate the general cubic graph to attain the second
graph, which has at least one turning point on the x-axis?

For example, given a graph of x^3 - 7x^2 + 15x - 11. If we translated this
graph using the vector (0, 2) -- so, 2 units in the positive direction of 
the y-axis, i.e., upwards -- we would get the equation 

   y = x^3 - 7x^2 + 15x - 9
     = (x - 1)(x - 3)^2

Here, the double root emerges; and therefore, the x-axis is a tangent to
the curve at x = 3, producing a minimum. Is the 'D' somehow tied into this
analogy? Is D the difference between the original intercept of the graph
(-11) and the translated graph (-9)? So would D (in this particular
example) be -11 - (-9) = -2?

But having said all that, Doctor Math ... if the above analogy is correct,
it was achieved through pure luck: I deliberately chose the equation 
(x - 1)(x - 3)^2, and just manipulated it. If we had such an equation as, 
say, x^3 - 7x^2 + 15x - 11, how would you know that translating it by 2 
units upwards would achieve a similar graph containing a double root? and
therefore a max/min point?

I'm really keen to understand if it is possible to calculate maximum and
minimum points of cubic graphs WITHOUT the use of calculus. However easy
it is, I want to explore this possibility.

I'd be most grateful if you would be able to please clarify 
this predicament.

Thanks, Doctor Math.

Date: 06/25/2016 at 19:31:26
From: Doctor Peterson
Subject: Re: Maximum and minimum of a cubic function WITHOUT calculus

> So with this said (daft as it may seem, but I may have answered my own
> question), is the 'D' the difference of the vertical component of the
> vector used to translate the general cubic graph to attain the second
> graph, which has at least one turning point on the x-axis?

Yes, that's exactly what I was saying. The new equation is the original
equation shifted D units downward, where we want D to be the y-coordinate
of the local extremum we are interested in. 

To make this clearer, here's a graph of the original equation:


                        |               *
                        |
                        |
                        |
                 *      |              *
             *        * |
                        |
           *            *            *
                        |
                        | *        *
    ......*.............|......*...............
          :             |      :        ^
          :             |      :        |
          :             |      :        |D
          :             |      :        |
          :             |      :        v
    -----*+-------------+------+---------------
          p             |      q
                        |
                        |
                        |


Let's try carrying out the whole process (below) so you can see that this
is exactly what my D is. You added 2 to your function to shift it upward,
changing both the y-intercept, and, more to the point, the y-coordinate of
the minimum we're interested in, bringing it to 0.

Note that if I were creating a problem with which to demonstrate the
method, I might have created the equation in just the way you did, first
writing a factored form so that I know that the extremum (or at least one
of them) has nice integer coordinates, and then modifying it so the
y-coordinate is not already zero. (Actually, I'd probably start with a
derivative with *two* nice zeros, in factored form, and integrate that to
get an appropriate equation to work on. Your way, we have no guarantee
that the maximum will also be a nice number.)

> I'm really keen to understand if it is possible to calculate maximum and
> minimum points of cubic graphs WITHOUT the use of calculus. However easy
> it is, I want to explore this possibility.

Let's do it. 

The original conversation, above, answers your question didactically,
showing how to find D ... eventually; but looking at it concretely would
help anyone fully grasp it.

We have the cubic equation

   y = x^3 - 7x^2 + 15x - 11

We translate this downward by an unknown distance D (the y-coordinate of
the extremum in question), with the goal of having a double zero:

   y = x^3 - 7x^2 + 15x - 11 - D

That is, we want, for some p and q,

   (x - p)(x - q)^2 == x^3 - 7x^2 + 15x - 11 - D

Here, "==" represents "is identically equal to." That is, they must be
equal for all x. Note that our "a" is 1.

By the way, another way to say this is that we want to take this ...

   x^3 - 7x^2 + 15x - 11 
   
... and rewrite it as this:

   (x - p)(x - q)^2 + D

We expand the left side, as I showed, resulting in

   x^3 - (p + 2q)x^2 + q(2p + q)x - pq^2
   == x^3 - 7x^2 + 15x - 11 - D

We conclude that we need

   -(p + 2q) = -7
   q(2p + q) = 15
       -pq^2 = -11 - D

We solve the first equation for p:

   p + 2q = 7
        p = 7 - 2q

We put this into the second equation:

   q(2(7 - 2q) + q) = 15
         q(14 - 3q) = 15
    3q^2 - 14q + 15 = 0

We solve this by factoring (since it has rational solutions, we don't need
to use the quadratic formula):

   (3q - 5)(q - 3) = 0
                 q = 5/3 or 3

These will be the x-coordinates of the two turning points. To find the
amount by which we have to translate to get x = 3, we put q = 3 and 
p = 7 - 2(3) = 1 into the last of our equations:

   -pq^2 = -11 - D
       D = pq^2 - 11 
         = (1)(3)^2 - 11 = -2

This tells us that our original equation ...

   y = x^3 - 7x^2 + 15x - 11

... can be written as

   y = (x - 1)(x - 3)^2 - 2

This last bit looks quite different from what I did in the general case,
because I don't have crowds of letters everywhere and could use a
different approach, which makes some parts feel more natural, but hides
the fact that D = aq^3 + bq^2 + cq + d.

Now, the OTHER extremum is found by taking q = 5/3, so that p = 11/3. Then
D turns out to be

   D = pq^2 - 11
     = (11/3)(5/3)^2 - 11 
     =  275/27 - 11 
     = -22/27

So we can also write our equation as

   y = (x - 11/3)(x - 5/3)^2 - 22/27

You can check this out to see that it is true.

I hope that helps! It was interesting revisiting this old discussion
and looking at it from a different direction.

- Doctor Peterson, The Math Forum at NCTM
  

Date: 06/27/2016 at 14:50:15
From: Yousuf
Subject: Thank you (Maximum and minimum of a cubic function WITHOUT calculus)

Hi, Doctor Peterson,

I want to say a HUUUGE thank you for helping clarify this article in the
most subtle way. You explained it in the example and it all clicked in
perfectly well. Now I can associate the points in the example seamlessly
with the algebraic proof -- all thanks to your help!!

Again I'm most thankful for your kind support, Doctor Peterson :)

- Yousuf.
Associated Topics:
High School Basic Algebra
High School Functions

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