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

We have:

$f(x) = x^2 + ax + b$ (a quadratic function)

$g(x) = f(x+1) - f(x-1)$ (the difference between f at two points)

$f(x) \geq 0$ for all real $x$ (f is always non-negative)

$g(20) = 72$

We need to find the smallest possible value of $b$.

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Let's substitute the definition of $f(x)$ into $g(x)$:

$g(x) = f(x+1) - f(x-1)$

First, let's find $f(x+1)$:

$f(x+1) = (x+1)^2 + a(x+1) + b$

Since $(x+1)^2 = x^2 + 2x + 1$:

$f(x+1) = x^2 + 2x + 1 + ax + a + b$

Next, let's find $f(x-1)$:

$f(x-1) = (x-1)^2 + a(x-1) + b$

Since $(x-1)^2 = x^2 - 2x + 1$:

$f(x-1) = x^2 - 2x + 1 + ax - a + b$

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Now we subtract:

$g(x) = f(x+1) - f(x-1)$

$g(x) = (x^2 + 2x + 1 + ax + a + b) - (x^2 - 2x + 1 + ax - a + b)$

Grouping like terms:

$x^2$ terms: $x^2 - x^2 = 0$

$x$ terms: $2x - (-2x) = 4x$

$ax$ terms: $ax - ax = 0$

Constant terms: $(1 + a + b) - (1 - a + b) = 2a$

Therefore: $g(x) = 4x + 2a$

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We're told that $g(20) = 72$. Let's substitute:

$g(20) = 4(20) + 2a = 80 + 2a$

Setting this equal to 72:

$80 + 2a = 72$

$2a = -8$

$a = -4$

So our function becomes: $f(x) = x^2 - 4x + b$

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A quadratic function is always non-negative if and only if it has no real roots or exactly one real root. If a quadratic has two distinct real roots, it would dip below the x-axis between those roots, making it negative somewhere.

For a quadratic $ax^2 + bx + c$ with $a > 0$, this happens when the discriminant $D = b^2 - 4ac \leq 0$.

The discriminant tells us about the nature of roots:

If $D > 0$: two distinct real roots (parabola crosses x-axis twice)

If $D = 0$: one real root (parabola touches x-axis once)

If $D < 0$: no real roots (parabola doesn't touch x-axis)

For our function $f(x) = x^2 - 4x + b$:

$a = 1$, coefficient of $x$ is $-4$, constant term is $b$

The discriminant is:

$D = (-4)^2 - 4(1)(b) = 16 - 4b$

For $f(x) \geq 0$ everywhere, we need:

$D \leq 0$

$16 - 4b \leq 0$

$16 \leq 4b$

$b \geq 4$

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The smallest possible value of $b$ is $4$.

At $b = 4$, our function becomes $f(x) = x^2 - 4x + 4 = (x-2)^2$, which is always non-negative and equals zero only at $x = 2$.