Question

expand 49+69x+42x^2+11x^3+x^4 in power of (x+2)​

Answer 1

Answer:

your ans is 2

hope it help

Answer 2

A Taylor series is a function's series expansion around a point.

We have the equation $49+69x+42{{x}^{2}}+11{{x}^{3}}+{{x}^{4}}$

Using Taylor's theorem, we shall extend the equation in the power of $(x+2)$.

Taylor's theorem is given as follows

$f(x)=f(a)+(x+a){f}''(a)+\frac{{{(x+a)}^{2}}}{2!}{f}''(a)+\frac{{{(x+a)}^{3}}}{3!}{f}'''(a)+\frac{{{(x+a)}^{4}}}{4!}{{{{f}'}'}'}'(a)$

Using $x+2=0$, we have

$f(x)={{x}^{4}}+11{{x}^{3}}+42{{x}^{2}}+69x+49$

$\begin{align} & f(-2)={{(-2)}^{4}}+11{{(-2)}^{3}}+42{{(-2)}^{2}}+69(-2)+49 \\ & f(-2)=16-88+168-138+49 \\ & \text{ }=7 \\ \end{align}$

$\begin{align} & {f}'(x)=4{{x}^{3}}+33{{x}^{2}}+84x+69 \\ & {f}'(-2)=-32+132-168+69 \\ & \text{ }=1 \\ \end{align}$

$\begin{align} & {f}''(x)=12{{x}^{2}}+66x+84 \\ & {f}''(-2)=12{{(-2)}^{2}}+66(-2)+84 \\ & \text{ }=0 \\ \end{align}$

$\begin{align} & {f}'''(x)=24x+66 \\ & {f}'''(-2)=24(-2)+66 \\ & \text{ }=18 \\ \end{align}$

$& f''''(x)=24 \\ & f''''(x)=0 \\ \end{align}$

Now we'll plug the numbers into Taylor's theorem equation.

$\begin{align} & f(x)=f(-2)+(x+2)f'(-2)+\frac{{{(x+2)}^{2}}}{2!}f''(-2)+\frac{{{(x+2)}^{3}}}{3!}f'''(-2)+\frac{{{(x+2)}^{4}}}{4!}f''''(-2) \\ & f(x)=7+(x+2)+\frac{{{(x+2)}^{2}}}{2!}0+\frac{{{(x+2)}^{3}}}{3!}18+\frac{{{(x+2)}^{4}}}{4!}24 \\ & f(x)=7+(x+2)+3{{(x+2)}^{3}}+{{(x+2)}^{4}} \\ \end{align}$

Hence, $49+69x+42{{x}^{2}}+11{{x}^{3}}+{{x}^{4}}$ can be expanded in the power of $(x+2)$ as

$f(x)=7+(x+2)+3{{(x+2)}^{3}}+{{(x+2)}^{4}}$