question 3 (1 point)\nsuppose f and g are functions with the following properties:\n\n$f(3)=g(3)=0,f(3)=f(3)=…
question 3 (1 point)\nsuppose f and g are functions with the following properties:\n\n$f(3)=g(3)=0,f(3)=f(3)=2,g(3)=g(3)=1.$\n\nif the second - degree taylor polynomial of $y = f(x)g(x)$ about the point $x = 3$ is given by\n\n$a_0 + a_1(x - 3)+a_2(x - 3)^2$\n\nthen $a_2$ is equal to which of the following?\n\n0\n2\n4\n-2\n-1
Answer
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 $x = 3$
We know that $f(3)=g(3)=0$, $f^\prime(3)=f^{\prime\prime}(3)=2$, and $g^\prime(3)=g^{\prime\prime}(3)=1$. Substitute $x = 3$ into the formula for $y^{\prime\prime}$: [ \begin{align*} y^{\prime\prime}(3)&=f^{\prime\prime}(3)g(3)+2f^\prime(3)g^\prime(3)+f(3)g^{\prime\prime}(3)\ &=2\times0 + 2\times2\times1+0\times1\ &=4 \end{align*} ]
Step3: Recall the formula for the Taylor polynomial
The second - degree Taylor polynomial of a function $y = h(x)$ about $x = a$ is $P_2(x)=h(a)+h^\prime(a)(x - a)+\frac{h^{\prime\prime}(a)}{2}(x - a)^2$. For $y = f(x)g(x)$ about $x = 3$, $a_2=\frac{y^{\prime\prime}(3)}{2}$. Since $y^{\prime\prime}(3)=4$, then $a_2 = 2$.
Answer:
2