Find cubic function whose graph has two horizontal tangents

Question

Find the cubic function $y=ax^3+bx^2+cx+d$ whose graph has horizontal tangents at the points $(−5,5)$ and $(3,-2)$.

I've seen this problem before with different numbers. For some reason I can't solve with these points. I've taken the derivative of $y=ax^3+bx^2+cx+d$ and then plugged in both x values. I set those two equations equal to zero. Then I plugged in the x values into $y=ax^3+bx^2+cx+d$. I then set the equations equal to $5$ and $-2$. So in the end I have four equations that I should be able to plug into a matrix. I ended up with $a = 7/121$, $b = 21/121$, $c = -315/121$, and $d = 325/121$. However, I'm being told that's incorrect.

Answer 1

First, the graph of that function passes through $(-5,5)$ and $(3,-2)$, that is, $$\begin{align} 5 = y(-5) &= -125a + 25b - 5c + d \\ -2 = y(3) &= 27a + 9b + 3c + d \end{align}$$ Also, since $y$ has critical points at $-5$ and $3$, we get $$\begin{align} 0 = y'(-5) &= 75a - 10b + c \\ 0 = y'(3) &= 27a + 6b + c \end{align}$$ Using this four equations we obtain that $$y(x) = \frac{7}{256}x^3 + \frac{21}{256}x^2 - \frac{315}{256}x + \frac{55}{256}$$

Answer 2

I get $$ \left( \begin{array}{rrrr|r} 27 & 6 & 1 &0 & 0 \\ 75 & -10 & 1 &0 & 0 \\ -125 & 25 & -5 &1 & 5 \\ 27 & 9 & 3 &1 & -2 \\ \end{array} \right) $$

Answer 3

Let $f(x)=ax^3+bx^2+cx+d$, then $f'(x)=3ax^2+2bx+c$

Since $f$ has horizontal tangents at $x=-5$ and $x=3$, then $$f'(x)=3a(x-(-5))(x-3)=3ax^2+6ax-45a$$ This is a parabola for which we know the two roots. Back to the original function $$f(x)=ax^3+3ax^2-45ax+d$$ We now have only two unknowns and we know two points of the cubic. Replacing the points in the function, we have $$5=f(-5)=-125a+75a+225a+d$$ $$-2=f(3)=27a+27a-135a+d$$ $$\begin{cases}5&=175a+d\\-2&=-81a+d\end{cases}$$ $$a=\frac{7}{256}\quad d=\frac{55}{256}$$ Finally $$f(x)=\frac{7}{256}x^3+\frac{21}{256}x^2-\frac{315}{256}x+\frac{55}{256}$$

Answer 4

There are other features of the cubic polynomial that can also be used to check the calculation of the coefficients. Since the two extrema (location of the horizontal tangent points) are given, with the larger function value at the smaller $ \ x-$coordinate, the function curve has an inflection, but is otherwise an increasing function; hence, $ \ a \ $ must be positive. The relative maximum and relative minimum are (anti-)symmetrically placed about the inflection point of a cubic polynomial (if they exist), which locates it in this case at $ \ \left(-1 \ , \ \frac32 \right) \ \ . \ $ We can then use the second derivative $ \ 6ax + 2b \ = \ 0 \ $ to establish that $ \ 6a·(-1) \ = \ -2b \ \Rightarrow \ b \ = \ 3a \ \ . \ $

The first derivative being $ \ 3ax^2 + 2bx + c \ \rightarrow \ 3ax^2 + 6ax + c \ = \ 3a·(x + 1 )^2 + (c - 3a) \ \ , \ $ the horizontal tangent point coordinates lead us to $$ 3a·(3 + 1)^2 \ + \ (c - 3a) \ \ = \ \ 0 \ \ \Rightarrow \ \ 3a·16 \ \ = \ \ 3a \ - \ c \ \ \Rightarrow \ \ c \ = \ -45a \ \ , $$ with the use of $ \ x \ = \ -5 \ $ producing the same relation. (So far the proportions of your coefficients are confirmed.)

We can now write our cubic polynomial as $ \ ax^3 + 3ax^2 - 45ax + d \ \ . \ $ The coordinates of the inflection point then tell us that $$ a·(-1)^3 \ + \ 3a·(-1)^2 \ - \ 45a·(-1) \ + \ d \ \ = \ \ -a \ + \ 3a \ + \ 45a \ + \ d \ \ = \ \ \frac32 \quad \mathbf{[ \ 1 \ ]} $$ $$ \Rightarrow \ \ d \ \ = \ \ \frac32 \ - \ 47a \ \ . $$

Returning to the relative extrema, which lie $ \ 4 \ $ units "to the left and right" of the inflection point and $ \ \frac72 \ $ units "above or below" it, we may write $$ a·(-1 \ \pm \ 4)^3 \ + \ 3a·(-1 \ \pm \ 4)^2 \ - \ 45a·(-1 \ \pm \ 4) \ + \ d \ \ = \ \ \frac32 \ \mp \ \frac72 \ \ , $$ or, subtracting equation $ \ \mathbf{1} \ $ above, $$ \begin{array}{c} a \ · \ (28 \ + \ 24 \ - \ 180) \ \ = \ \ -\frac72 \\ a \ · \ (-124 \ + \ 72 \ + \ 180) \ \ = \ \ +\frac72 \end{array} \ \ \ \Rightarrow \ \ \ \mp \ 128·a \ \ = \ \ \mp \ \frac72 \ \ \ \Rightarrow \ \ a \ \ = \ \ \frac{7}{256} \ \ . $$ [You didn't show your particular numerical calculations, but it appears the error was in determining this denominator.] With $ \ d \ = \ \frac32 \ - \ 47·\frac{7}{256} \ = \ \frac{3·128 \ - \ 7·47}{256} \ = \ \frac{55}{256} \ \ , \ $ we confirm the coefficients obtained in the other posted answers. (Indeed, $$ \frac{7}{256}·[ \ (-5)^3 \ + \ 3·(-5)^2 \ - \ 45·(-5) \ ] \ + \ \frac{55}{256} \ \ = \ \ \frac{7· 175 \ + \ 55}{256} \ \ = \ \ \frac{1280}{256} \ \ = \ \ +5 $$ and $$ \frac{7}{256}·[ \ 3^3 \ + \ 3·3^2 \ - \ 45·3 \ ] \ + \ \frac{55}{256} \ \ = \ \ \frac{7· (-81) \ + \ 55}{256} \ \ = \ \ \frac{-512}{256} \ \ = \ \ -2 \ \ . \ ) $$