Find a Cubic Function given an inflection point, critical point, and function value.

Question

Find a cubic function f(x) = ax^3 + bx^2 + cx + d

Given:

• Inflection point (0,18)
• Critical point x = 2
• F(2) = 2

I know how to solve for the general forms of the derivatives, and to set the values of the functions and the derivatives at those points, but the system of equations that I come up with lead me to the wrong answer.

My work so far is as follows:

• f(x) = ax^3 + bx^2 + cx + d
• f'(x) = 3ax^2 + 2bx + c
• f''(x) = 6ax + 2b

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• f(2) = a(2)^3 + b(2)^2 + c(2) + d = 2
• f(2) = 8a + 4b + 2c + d =2
• f'(2) = 3a(2)^2 + 2b(2) + c = 0
• f'(2) = 12a + 4b +c = 0
• f''(18) = 6a(18) + 2b = 0
• f''(18) = 108a + 2b = 0

Thus:

• f(2) = 8a + 4b + 2c + d =2
• f'(2) = 12a + 4b +c = 0
• f''(18) = 108a + 2b = 0

But here I get stuck and unable to solve for any particular variable.

Answer 1

An inflection point at (0,18) gives two equations:

$f''(0) = 0$ and $f(0)=18$.

The critical point gives rise to the equation $f'(2)=0$ and you have $f(2)=2$. Then you have four linear equations in four unknowns, which you can solve by substitution.

Answer 2

There is a useful theorem concerning cubic polynomial functions: the $ \ x-$ coordinate of the point of inflection is midway between the $ \ x-$ coordinates of the local extrema (if they exist). It is the analogue of the proposition for quadratic functions that the $ x-$ coordinate of the absolute extremum is midway between the zeroes (if they exist). So with the point of inflection given to be on the $ \ y-$axis and one critical point located at $ \ x = 2 \ \ , $ the other critical point is found at $ \ x = -2 \ \ . $

A significant implication is introduced with the inflection point located on the $ \ y-$axis. If we (temporarily) shift the graph of the function so that the inflection point is at the origin, the cubic function attains odd symmetry. This is seen in the symmetrical location of the critical points about the $ \ y-$axis. This will also mean that since the critical point at $ \ (2 \ , \ 2 ) \ $ is $ \ 16 \ $ units "below" the inflection point, the other critical point is $ \ 16 \ $ units "above" it. With the inflection point being the $ \ y-$intercept of the function curve $ \ (0 \ , \ \mathbf{d=18}) \ $ , we can say that the local minimum lies at $ \ (2 \ , \ d-16 = 2) \ \ $ and the local maximum at $ \ (-2 \ , \ d+16 = 34) \ \ . $

Note that your equation for the second derivative at the inflection point gives $ \ 6a·0 \ = \ 2b \ \Rightarrow \ \mathbf{b = 0} \ \ . $ This fits with our temporary shifting of the graph discussed above: having $ \ b = 0 \ $ makes the sum of the first three terms of the cubic polynomial an odd-symmetry function, $ \ ax^3 + cx \ \ . $

Incidentally, now that we know the relative locations of the local maximum and local minimum, with the maximum "to the left" of the minimum, it must be the case that $ \ a > 0 \ \ . $ We will resolve this fully in a moment.

Your quadratic polynomial for the first derivative can be expressed as $$ 3ax^2 \ + \ 2bx \ + \ c \ \ = \ \ 3a · (x+2) · (x-2) \ \ = \ \ 3ax^2 \ - \ 12a \ \ , $$ again confirming that $ \ b = 0 \ \ . $ We can also immediately conclude that $ \ c = -12a \ \ . $

We are nearly finished: the cubic polynomial has the form $ \ f(x) \ = \ ax^3 - 12ax + 18 \ , \ a > 0 \ \ . $ We may use either of our known critical points to find the remaining coefficient: $$ f(2) \ \ = \ \ a·2^3 \ - \ 12a·2 \ + \ 18 \ \ = \ \ -16a \ + \ 18 \ \ = \ \ 2 \ \ \Rightarrow \ \ \mathbf{a = 1} \ \ \ \text{or} $$ $$ f(-2) \ \ = \ \ a·(-2)^3 \ - \ 12a·(-2) \ + \ 18 \ \ = \ \ 16a \ + \ 18 \ \ = \ \ 34 \ \ \Rightarrow \ \ a = 1 \ \ . $$ (Hence, $ \ \mathbf{c} = \ -12·1 \ \ \mathbf{ = -12} \ \ . ) $

Our cubic polynomial function is therefore $ \ f(x) \ = \ x^3 \ - \ 12x \ + \ 18 \ \ , $ for which a graph is presented below.