1

$$f(x) = a x^3 + b x^2 + c x + d$$

The graph of $f$ definitely has a point of inflexion. Which of the coefficients $a, b, c, d$ determines that the point of inflexion of the function $f$ lies on the vertical axis? Write down this coefficient and the corresponding condition."

I tried to calculate it, the answer doesn't seem to me that simple. the second derivative is:

$$f''(x)=6ax+2b$$

so the coordinates of the point of inflexion are $(-b/(3a),f(-b/(3a))$ And I know that $f(b/(3a))=0$, because it lies on the $x$ axis. But this is one equation with four variables. Where did I overcomplicate it?

Thank you

Robert Lewis
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TTomi
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1 Answers1

1

With

$f(x) = ax^3 + bx^2 + cx + d, \tag 1$

we have

$f'(x) = 3ax^2 + 2bx + c, \tag 2$

and

$f''(x) = 6ax + 2b; \tag 3$

points of inflection occur at those $x$ for which

$f'(x) = f''(x) = 0, \tag 4$

that is, where

$6ax + 2b = 0, \tag 5$

and $3ax^2 + 2bx + c = 0; \tag 6$

(5) is possessed of the unique solution

$x = -\dfrac{b}{3a}, \tag 7$

if there is an inflection point, this $x$ must also satisfy (6), whence

$\dfrac{b^2}{3a} - \dfrac{2b^2}{3a} + c = c - \dfrac{b^2}{3a} = 0. \tag 8$

Now I assume

$a \ne 0, \tag 9$

necessary for $f(x)$ to be a bona fide cubic; thus the expressions (7)-(8) make sense; for the inflection point to lie on the vertical axis $(0, y)$ we must have

$x = -\dfrac{b}{3a} = 0 \Longrightarrow b = 0,\tag{10}$

and then via (8)

$c = 0 \tag{11}$

as well.

It thus appears that $b$ is the coefficient of (1) determinative of whether an inflection point lies on the vertical axis; we observe that $b$ also specifies $c$ via (8), so that (1) becomes

$f(x) = ax^3 + d, \tag{13}$

a simple cubic translated up or down in $y$ according to $d$.

Robert Lewis
  • 71,180