Ever seen that TV game show where a piano player plays just a few notes of a tune and the MC then exhorts a contestant to "Name That Tune." They do give the contestant a few hints -- perhaps a brief description of the show the tune is from or when the tune was popular. But each contestant has to rely mostly on his or her personal store of musical knowledge.

Some calculus teachers will do the same to you. They
will give you a little information about a function --
that it is a cubic, say -- and then play a few of its notes.
And "notes" for a function will be values at some specified
`x`'s of what the function or its first or second
derivatives are. Then for the grand prize of full credit,
you will have to Name That Function.

It's not as hard as it sounds. The contestants on the TV show had their personal stores of musical knowledge. And you have your personal store of algebra and calculus knowledge to draw on. You will need to use only a small portion of it to solve one of these.

Also, you never are asked to solve for the whole function. They always ask you for just a few coefficients. Those few coefficients tell you what the whole function is. Here's an example:

Find `A`, `B`, and `C` if the function

f(x) = xhas an inflection point at^{3}+ Ax^{2}+ Bx + C eq. 8.3-1

Recall that a *critical point* or *stationary point*
is a property of the first derivative of a function, namely it
is an `x` where the first derivative is zero. Likewise
an *inflection point* is a property of the second derivative
of a function, namely it is an `x`
where the second derivative is equal to
zero. So it only makes sense that the first thing to do with
this problem is to find the first and second derivatives of
`f(x)`.

f'(x) = 3xThe statement that there is a critical point at^{2}+ 2Ax + B eq. 8.3-2a f"(x) = 6x + 2A eq. 8.3-2b

0 = 3(1/9) + 2A(1/3) + B eq. 8.3-3aThe statement that there is an inflection point at

0 = 6(-17/6) + 2A eq. 8.3-3bThis last equation is a snap to solve for

0 = 3(1/9) + 17/3 + B = 18/3 + B eq. 8.3-4You should be able to solve that one quickly to find that

So what about `C`? That's the easiest one. When the problem tells you
that the `y` intercept is `3`, it is telling you that
`f(0) = 3``x`
and `3` in for `f(x)`:

3 = C

In the problem above we were able to find the coefficients one at a time.
Sometimes that is not possible. In those cases you have to find them
*simultaneously*. Here's an example:

Find `A`, `B`, and `C` if the function,

f(x) = xif^{3}+ Ax^{2}+ Bx + C eq. 8.3-5

Since the problem talks about critical points, and critical points
are a property of the first derivative, clearly we need write
the expression for `f'(x)`

f'(x) = 3x^{2}+ 2Ax + B eq. 8.3-6

The first clue tells you of a critical point at `x = -2/3``x`, you have `f'(x) = 0``f'(x)` with zero and replace `x` with `-2/3`:

0 = 3(4/9) + 2A(-2/3) + B = 4/3 - (4/3)A + B eq. 8.3-7aThe second clue tells you of a critical point at

0 = 3(25) + 2A(5) + B = 75 + 10A + B eq. 8.3-7bThe last clue tells you to take

3/2 = 1 + A + B + C eq. 8.3-7cYou can solve for

0 = 225/3 + (30/3)A + B - 0 = 4/3 - ( 4/3)A + BNow plug this value for~~0 = 221/3 + (34/3)A A = -13/2~~

0 = 75 - 65 + B B = -10Finally plug the values for both

3/2 = 1 - 13/2 - 10 + C C = 17

It's not just polynomials that you can play this game with. It works for lots of other kinds of functions as well:

Find `A` and `k` if the function

f(x) = Ax ehas a critical point at^{-kx}eq. 8.3-8

Once again the problem asks about a critical point, so once again
you must find the first derivative. Since `f(x)` is a product,
you have to employ the

f'(x) = Ae^{-kx}- Akx e^{-kx}= A(1 - kx)e^{-kx}eq. 8.3-9

The clue tells of a critical point at `x = 10``f'(x)`, and put in
`10` for `x`:

0 = A(1 - 10k)eWhen you divide^{-10k}eq. 8.3-10

Once you know what `k` is, it's easy to substitute it back into
`f(x)` and determine `A`.

f(x) = Ax eThe second clue says that at the critical point, which is at^{-0.1x}eq. 8.3-11

12 = 10Ae^{-1}= 10A/e eq. 8.3-12a A = 1.2e = 3.261938194... eq. 8.3-12b

What if you had to do the last problem again but this time, instead of the clue telling you about a critical point, it told you about an inflection point of the same function:

Find `A` and `k` if the function

f(x) = Ax ehas an inflection point at^{-kx}eq. 8.3-8

So now, instead of the first derivative, you have to examine the second derivative. You already know from equation 8.3-9 that

f'(x) = AeYou apply the product rule to^{-kx}- Akx e^{-kx}= A(1 - kx)e^{-kx}eq. 8.3-9

f"(x) = -AkeThe clue tells of an inflection point at^{-kx}- Ak(1 - kx)e^{-kx}= A(k^{2}x - 2k)e^{-kx}eq. 8.3-13

0 = A(10kWhen you divide out the^{2}- 2k)e^{-10k}eq. 8.3-14

0 = 10kSo this time you get^{2}- 2k eq. 8.3-15a 0 = 10k - 2 eq. 8.3-15b

Again, solving for `A` is a piece of cake once you know `k`. The second
clue says that at the inflection point the function is equal to `12`. So take
the original function, put in `0.2` for `k`, `10` for
`x`, and `12` for `f(x)`:

12 = 10Ae^{-2}eq. 8.3-16a A = 1.2e^{2}= 8.866867319... eq. 8.3-16b

These two problems might look, at first, to be completely different (and much harder) than the worked examples above, but they're not. Simply use the same steps of setting the function and its derivative to zero at the values named in the problem, and then solve for the coefficients.

**1)** Find `l` and
`w` if the function,

f(t) = ehas a zero crossing at^{lt}cos(wt)

**2)** Find `l` and `A` if the function,

f(t) = ehas a zero crossing at^{lt}(sin(t) + A cos(t))

Suppose they tell you it's a polynomial and just give you a
handful of points that it must pass through. This does not
require calculus to solve, just some algebra. So you can
consider this problem to be **optional material**, but
it is useful to be able to solve this type of problem.
Here's a sample:

Find the cubic (3rd degree) polynomial, `f(x)`, such that

f(1) = -1 f(-1) = -17 f(2) = 19 f(-2) = -49 Table 8.3-1One way to do this one is to say that

f(x) = AxNow you try to solve for^{3}+ Bx^{2}+ Cx + D eq. 8.3-17

f(1) = -1 = A + B + C + D eq. 8.3-18aNow put in

f(-1) = -17 = -A + B - C + D eq. 8.3-18bNow you do the same with

f(2) = 19 = 8A + 4B + 2C + D eq. 8.3-18c f(-2) = -49 = -8A + 4B - 2C + D eq. 8.3-18dYou would then have to solve these four equations simultaneously for

Well there is a way.
Table 8.3-1 shows that we are considering `x`'s at
`1`, `-1`, `2`, and `-2`. So
you form the following expression:

(x - 1)(x + 1)(x - 2)(x + 2) eq. 8.3-19Each factor is

(x + 1)(x - 2)(x + 2) eq. 8.3-20aThis expression is zero at

(x - 1)(x - 2)(x + 2) eq. 8.3-20bis zero at

(x + 1)(x - 1)(x + 2) eq. 8.3-20c (x + 1)(x - 1)(x - 2) eq. 8.3-20dEach of those four expressions is a 3rd degree polynomial. If you multiply them out you get

PSuppose that you imagine_{1}(x) = (x + 1)(x - 2)(x + 2) = x^{3}+ x^{2}- 4x - 4 eq. 8.3-21a P_{2}(x) = (x - 1)(x - 2)(x + 2) = x^{3}- x^{2}- 4x + 4 eq. 8.3-21b P_{3}(x) = (x + 1)(x - 1)(x + 2) = x^{3}+ 2x^{2}- x - 2 eq. 8.3-21c P_{4}(x) = (x + 1)(x - 1)(x - 2) = x^{3}- 2x^{2}- x + 2 eq. 8.3-21d

f(x) = kPIf you solve for_{1}(x) + lP_{2}(x) + mP_{3}(x) + nP_{4}(x) eq. 8.3-22

Here you can solve for each of
`k`,
`l`,
`m`, and
`n` individually.
You already know that at ` x = 1`

POnly_{2}(1) = P_{3}(1) = P_{4}(1) = 0

f(1) = kPYou can easily find that_{1}(1) = -1 eq. 8.3-23a

1 2 k =~~=~~~~6 12~~

By using the other `x` values in

PYou find that_{1}(-1) = P_{3}(-1) = P_{4}(-1) = 0 P_{1}(2) = P_{2}(2) = P_{4}(2) = 0 P_{1}(-2) = P_{2}(-2) = P_{3}(-2) = 0

f(-1) = lPFrom those equations you can find that_{2}(-1) = -17 eq. 8.3-23b f(2) = mP_{3}(2) = 19 eq. 8.3-23c f(-2) = nP_{4}(-2) = -49 eq. 8.3-23d

17 34 l = -Now to find the coefficients of the polynomial for~~= -~~~~6 12 19 m =~~~~12 49 n =~~~~12~~

f(x) = (2/12)PIf you plug all the_{1}(x) + (-34/12)P_{2}(x) + (19/12)P_{3}(x) + (49/12)P_{4}(x) = (2/12) ( x^{3}+ x^{2}- 4x - 4) + (-34/12)( x^{3}- x^{2}- 4x - 4) + (19/12) ( x^{3}+ 2x^{2}- x - 2) + (49/12) ( x^{3}- 2x^{2}- x + 2) = (1/12) ( 2x^{3}+ 2x^{2}- 8x - 8) + (1/12) (-34x^{3}+ 34x^{2}+ 136x - 136) + (1/12) ( 19x^{3}+ 38x^{2}- 19x - 38) + (1/12) ( 49x^{3}- 98x^{2}- 49x + 98) =~~(1/12) ( 36x~~^{3}- 24x^{2}+ 60x - 84) = 3x^{3}- 2x^{2}+ 5x - 7 eq. 8.3-24

One final note. If you have an `n`th degree polynomial,
it can be made to pass exactly through only `n+1` specified points.
In this problem we had 4 points, so the polynomial had to be
a minimum of 3rd degree, that is a cubic. A polynomial that is made to
pass exactly through 3 points would have to be a quadratic, and
one and only one quadratic will pass through any 3 given points.
A polynomial that is made to pass through exactly 2 points,
well you already knew that it had to be a straight line, which
is a first degree polynomial.

Move on to Little Red Riding Hood Goes to Town (approximation and intro to Taylor Series)

email me at *hahn@netsrq.com*