calculate the taylor polynomials t2 and t3 centered at x = a for the function f(x) = 5ln(x + 1), a = 0…

Explanation:

Step1: Recall Taylor - polynomial formula

The Taylor polynomial of degree $n$ centered at $x = a$ for a function $f(x)$ is given by $T_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k$, where $f^{(k)}(x)$ is the $k$-th derivative of $f(x)$. Here $a = 0$, so $T_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(0)}{k!}x^k$. First, find the derivatives of $f(x)=5\ln(x + 1)$. The first - derivative: $f^{\prime}(x)=\frac{5}{x + 1}$. The second - derivative: $f^{\prime\prime}(x)=-\frac{5}{(x + 1)^2}$. The third - derivative: $f^{\prime\prime\prime}(x)=\frac{10}{(x + 1)^3}$. And $f(0)=5\ln(0 + 1)=0$, $f^{\prime}(0)=\frac{5}{0 + 1}=5$, $f^{\prime\prime}(0)=-\frac{5}{(0 + 1)^2}=-5$, $f^{\prime\prime\prime}(0)=\frac{10}{(0 + 1)^3}=10$.

Step2: Calculate $T_2(x)$

For $n = 2$, $T_2(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^2$. Substitute $f(0) = 0$, $f^{\prime}(0)=5$, $f^{\prime\prime}(0)=-5$ into the formula: $T_2(x)=0 + 5x+\frac{-5}{2}x^2=5x-\frac{5}{2}x^2$.

Step3: Calculate $T_3(x)$

For $n = 3$, $T_3(x)=f(0)+f^{\prime}(0)x+\frac{f^{\prime\prime}(0)}{2!}x^2+\frac{f^{\prime\prime\prime}(0)}{3!}x^3$. Substitute $f(0) = 0$, $f^{\prime}(0)=5$, $f^{\prime\prime}(0)=-5$, $f^{\prime\prime\prime}(0)=10$ into the formula: $T_3(x)=0 + 5x+\frac{-5}{2}x^2+\frac{10}{6}x^3=5x-\frac{5}{2}x^2+\frac{5}{3}x^3$.

Answer:

$T_2(x)=5x-\frac{5}{2}x^2$ $T_3(x)=5x-\frac{5}{2}x^2+\frac{5}{3}x^3$