C Roettger, S15 – Math 165, Project 3A
You are Function Detectives! Evil Genius Dr Moriarty’s secret is a function f (x), about which you have discovered the following clues.
1. It is a polynomial.
2. It has inflection points at x = −2, 0, 2.
√
3. It has horizontal tangent lines at x = ±1 and at x = ± 7.
4. Its graph passes through the point (0, 1).
5. Its graph passes through the point (1, 3).
Your ultimate goal is to find the function f (x) itself. Fortunately, the clues
are already in the right order to solve the puzzle. In particular, use 2. to
determine the degree of f (x). Then use it to determine f �� (x). Use antiderivatives and 3. to examine f � (x). This leaves you with two parameters
undetermined. The last two clues nail these down. Have some fun by plotting several (eg five) different functions satisfying all clues 1. - 4., including
the correct one. Include the plot in your writeup.
Solution. From clue # 2, we know that f �� (x) = 0 for x = −2, 0, 2
(actually, f �� (x) has to change sign there). So f (x) has factors x + 2 and x
and x − 2. In total,
f �� (x) = A(x + 2)x(x − 2) = Ax(x2 − 4) = A(x3 − 4x).
Let us find f � (x) as an antiderivative of f �� (x),
�
� 4
�
x
2
�
3
− 2x + C .
f (x) =
A(x − 4x) dx = A
4
(it is more convenient if we define C by being inside the parenthesis than
�
outside, but it could be
√ done either way). Next, clue # 3 says f2 (x) = 0
for x = ±1 and x = ± 7. These conditions happen exactly for x = 1 and
x2 = 7. So, putting x2 = 1,
1
−2+C =0
4
which means C = 74 . Then automatically, f � (x) = 0 for x2 = 7 as well. We
still do not know the value of A. Even clue # 4 does not determine its value.
But since
� 4
�
�
x
7
2
f (x) =
A
− 2x +
dx
4
4
�
�
A x5 8x3
=
−
+ 7x + D
4 5
3
the clue #4 tells us D = 1. Finally, clue # 5 says
�
�
A 1 8
f (1) =
− +7 +1=3
4 5 3
which we can solve for A to get
A=
8 · 15
30
= .
68
17
The graph below shows several functions that satisfy all clues # 1 - 4, with
the one plotted in bold red being Dr Moriarty’s secret. Note how they all
share the x-coordinates of points with horizontal tangent lines (these are local
maxima / minima). It’s harder to see that they all have inflection points with
x-coordinates −2 and 2 because all the graphs meet in a different point close
by. But the common inflection point (0, 1) is very visible.