suppose f and g are functions with the following properties: f(2) = g(2) = 0, f(2) = g(2) = 2, f(2) = 1…

Explanation:

Step1: Recall product - rule for derivatives

The first - derivative of $y = f(x)g(x)$ using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$ is $y^\prime=f^\prime(x)g(x)+f(x)g^\prime(x)$. The second - derivative of $y = f(x)g(x)$ using the product - rule on $y^\prime$: [ \begin{align*} y^{\prime\prime}&=(f^{\prime\prime}(x)g(x)+f^\prime(x)g^\prime(x))+(f^\prime(x)g^\prime(x)+f(x)g^{\prime\prime}(x))\ &=f^{\prime\prime}(x)g(x)+2f^\prime(x)g^\prime(x)+f(x)g^{\prime\prime}(x) \end{align*} ]

Step2: Evaluate $y^{\prime\prime}$ at $x = 2$

We know that $f(2)=g(2)=0$, $f^\prime(2)=g^\prime(2)=2$, $f^{\prime\prime}(2)=1$, and $g^{\prime\prime}(2)=-2$. Substitute these values into the formula for $y^{\prime\prime}(x)$: [ \begin{align*} y^{\prime\prime}(2)&=f^{\prime\prime}(2)g(2)+2f^\prime(2)g^\prime(2)+f(2)g^{\prime\prime}(2)\ &=1\times0 + 2\times2\times2+0\times(-2)\ &=8 \end{align*} ]

Step3: Recall the formula for the coefficient of the second - degree term in Taylor polynomial

The second - degree Taylor polynomial of a function $y = h(x)$ about $x = a$ is $h(a)+h^\prime(a)(x - a)+\frac{h^{\prime\prime}(a)}{2!}(x - a)^2+\cdots$. For $y = f(x)g(x)$ about $x = 2$, the coefficient $a_2=\frac{y^{\prime\prime}(2)}{2!}$. Since $y^{\prime\prime}(2)=8$, then $a_2=\frac{8}{2}=4$.

Answer:

C. 4