Question

Prove that every quadratic function has exactly one local extremum.

Short answer

It is proded that every quadratic function has exactly one local extremum.

Step-by-step solution

Step 1. Given Information 

We are given that every quadratic function has exactly one local extremum.

Step 2. Proving the statement 

Consider the quadratic function,

$f\left(x\right)=a{x}^2+bx+c\text{;where}a\ne 0\text{.}$

The objective is to prove that f has exactly one local extremum. The first derivative of the function is given by,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{d}{dx}\left(a{x}^2+bx+c\right) \\ =2ax+b \end{gathered}$$

For critical values,

role="math" localid="1648437135068" $$\begin{gathered} f\text{'}\left(x\right)=0 \\ 2ax+b=0 \\ 2ax=-b \\ x=-\frac{b}{2a} \end{gathered}$$

As x has only one real value, The function has exactly one local extremum. Therefore, every quadratic function has exactly one local extremum.