Plug your expressions for $f(x+2)$, $f(x+1)$, and $f(x)$ into $g(x)=f(x+2)-4f(x+1)+4f(x)$. Look for a common factor you can factor out.

After factoring, you should have $g(x)$ as a product of $ab^x$ and a quadratic in $b$. Since $a\neq 0$ and $b>0$, which factor must be zero for $g(x)$ to be zero for every real $x$?

Set the quadratic factor equal to zero and solve for $b$. Check that your solution also satisfies the conditions $b>0$ and $b\neq 1$.

From the algebraic work, $g(x)$ simplifies to $ab^x(b^2-4b+4)$. For checking on Desmos, you only need to examine the factor $b^2-4b+4$ (you can ignore $a$ and $b^x$ since they are never zero for the allowed values).

In Desmos, type the expression b^2 - 4b + 4. Then, for each answer choice, replace b with that value (for example, (1/2)^2 - 4*(1/2) + 4, (sqrt(2))^2 - 4*sqrt(2) + 4, 2^2 - 4*2 + 4, and 4^2 - 4*4 + 4).

Look at the numerical result for each substitution. The correct value of $b$ is the one that makes the expression b^2 - 4b + 4 equal to 0, since that would make $g(x)=ab^x(b^2-4b+4)$ equal to 0 for all real $x$.

We are given $f(x)=ab^x$.

Use this to write the shifted inputs:

• $f(x+1)=ab^{x+1}=ab^x\cdot b$
• $f(x+2)=ab^{x+2}=ab^x\cdot b^2$

Keep $f(x)=ab^x$ as is. We will substitute these into $g(x)$ next.

Now substitute the expressions from Step 1 into

We get

Factor out the common factor $ab^x$:

So $g(x)$ is a product of $ab^x$ and the quadratic expression $b^2-4b+4$.

We are told $a\neq 0$, $b>0$, and $b\neq 1$. For any such $a$ and $b$, the factor $ab^x$ is never $0$ for any real $x$.

Therefore, the only way $g(x)=ab^x\bigl(b^2-4b+4\bigr)$ can equal $0$ for all real $x$ is if the other factor is zero:

Factor this quadratic:

Set it equal to $0$:

This value satisfies $b>0$ and $b\neq 1$, so the correct choice is $\boxed{2}$ (answer C).