Let $f\left(x\right)=x^2+ax+b$ and $g\left(x\right)=f(x+1)-f(x-1)$. If $f\left(x\right)\ge 0$ for all real x, and $g\left(20\right)=72$, then the smallest possible value of b is

The correct option is 4.

To find the smallest possible value of $b$, we can break the problem down into the following steps:

Step 1: Simplify the expression for $g\left(x\right)$
We are given the function $f\left(x\right)=x^2+ax+b$.
Let us evaluate $f(x+1)$ and $f(x-1)$:
$f(x+1)={(x+1)}^2+a(x+1)+b$
$f(x-1)={(x-1)}^2+a(x-1)+b$

Now, we substitute these into the definition of $g\left(x\right)=f(x+1)-f(x-1)$:
$g\left(x\right)=[{(x+1)}^2+a(x+1)+b]-[{(x-1)}^2+a(x-1)+b]$

Using the difference of squares and expanding the terms, we get:
${(x+1)}^2-{(x-1)}^2=4x$
$a(x+1)-a(x-1)=2a$
$b-b=0$

Adding these parts together gives us:
$g\left(x\right)=4x+2a$

Step 2: Find the value of $a$
We are given that $g\left(20\right)=72$. Substituting $x=20$ into our simplified equation for $g\left(x\right)$:
$4\left(20\right)+2a=72$
$80+2a=72$
$2a=-8$
$a=-4$

Step 3: Apply the condition $f\left(x\right)\ge 0$
We are given that $f\left(x\right)=x^2+ax+b\ge 0$ for all real $x$.
For a quadratic equation with a positive leading coefficient to be non-negative for all real values of $x$, its discriminant must be less than or equal to zero:
$D=a^2-4\left(1\right)\left(b\right)\le 0$
$a^2-4b\le 0$
$b\ge \frac{a^2}{4}$

Substituting the value of $a=-4$ into this inequality:
$b\ge \frac{{(-4)}^2}{4}$
$b\ge \frac{16}{4}$
$b\ge 4$

Therefore, the smallest possible value of $b$ is $4$.