Question

Prove that every cubic function (i.e., every function of the form $f\left(x\right)=a{x}^3+b{x}^2+cx+d$for some constants a, b, c, and d) has exactly one inflection point. (Note: It is not enough just to show that the second derivative of any cubic function has exactly one zero; you must also show that the sign of the second derivative changes.)

Short answer

Every cubic function has exactly one inflection point which occur at$x=-\frac{b}{3a}.$

Step-by-step solution

Step 1. Given Information.

Given a cubic function:$f\left(x\right)=a{x}^3+b{x}^2+cx+d$.

Step 2. Theorem.

The Derivative Measures Where a Function is Increasing or Decreasing

Let f be a function that is differentiable on an interval I.

(a) If $f\text{'}$is positive in the interior of I, then f is increasing on I.

(b) If $f\text{'}$is negative in the interior of I, then f is decreasing on I.

(c) If $f\text{'}$is zero in the interior of I, then f is constant on I.

Step 3. Proof.

$$\begin{gathered} f\left(x\right)=a{x}^3+b{x}^2+cx+d \\ Differentiatingbothsidesw.r.txweget, \\ f\text{'}\left(x\right)=3a{x}^2+2bx+c. \end{gathered}$$

Inflection point is the point where the value of second derivative of the function is zero. So,

$$\begin{gathered} f\text{'}\left(x\right)=3a{x}^2+2bx+c, \\ Differentiatingagainw.r.txweget, \\ f\text{'}\text{'}\left(x\right)=6ax+2b, \\ and, \\ f\text{'}\text{'}\left(x\right)=0thismeans, \\ 6ax+2b=0, \\ and,x=-\frac{2b}{6a}=-\frac{b}{3a}. \\ So,wegetthatthefunctionwillhavesurelyoneinflectionpointwhichoccuratx=-\frac{b}{3a}. \end{gathered}$$