Q.If $f(x) = x^2 + 3x$ and $h = 1$ then show that $\Delta f(x) = 2x + 4$

With $h=1$, $\Delta f(x)=f(x+1)-f(x)$; expanding $f(x)=x^2+3x$ gives $2x+4$.

The forward difference operator with step $h=1$ is $\Delta f(x)=f(x+h)-f(x)=f(x+1)-f(x)$.

Compute $f(x+1)$ for $f(x)=x^2+3x$:

$f(x+1) = (x+1)^2 + 3(x+1) = x^2 + 2x + 1 + 3x + 3 = x^2 + 5x + 4.$

Now subtract $f(x)=x^2+3x$:

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