Discussion
The value of \(\Delta^{2} f(x)\), where \(f(x) = ab^{cx}\) is :
Answer: A
\(\Delta^{2} f(x)\) equals \((b^{ch} - 1)^{2}\, ab^{cx}\).
With step h:
\[ \begin{aligned} \Delta f(x) &= ab^{c(x + h)} - ab^{cx} = (b^{ch} - 1)\, ab^{cx} \\ \Delta^{2} f(x) &= (b^{ch} - 1)\,\Delta\left(ab^{cx}\right) = (b^{ch} - 1)^{2}\, ab^{cx} \end{aligned} \]
Each application multiplies by the same factor \(b^{ch} - 1\), because an exponential is an eigenfunction of \(\Delta\).
The options with \(b^{ch} + 1\) have the wrong sign (they would come from a sum, not a difference), and \((ab^{cx})^{2}\) squares the function instead of applying \(\Delta\) twice.
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