question 9 of 10 find the taylor polynomial t₂(x) and compute the error |f(x) - t₂(x)| for the given values…

Explanation:

Step1: Recall Taylor - polynomial formula

The second - degree Taylor polynomial $T_2(x)$ of a function $y = f(x)$ centered at $a$ is given by $T_2(x)=f(a)+f^{\prime}(a)(x - a)+\frac{f^{\prime\prime}(a)}{2!}(x - a)^2$. First, find the derivatives of $f(x)=e^{\sin(x)}$. Using the chain - rule, if $y = e^u$ and $u=\sin(x)$, then $f^{\prime}(x)=e^{\sin(x)}\cos(x)$. For the second - derivative, using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = e^{\sin(x)}$ and $v=\cos(x)$. So $f^{\prime\prime}(x)=e^{\sin(x)}\cos^2(x)-e^{\sin(x)}\sin(x)=e^{\sin(x)}(\cos^2(x)-\sin(x))$.

Step2: Evaluate $f(a)$, $f^{\prime}(a)$ and $f^{\prime\prime}(a)$

Given $a=\frac{\pi}{2}$, then $f(\frac{\pi}{2})=e^{\sin(\frac{\pi}{2})}=e^1 = e$, $f^{\prime}(\frac{\pi}{2})=e^{\sin(\frac{\pi}{2})}\cos(\frac{\pi}{2})=e\times0 = 0$, $f^{\prime\prime}(\frac{\pi}{2})=e^{\sin(\frac{\pi}{2})}(\cos^2(\frac{\pi}{2})-\sin(\frac{\pi}{2}))=e(0 - 1)=-e$.

Step3: Find $T_2(x)$

Substitute into the Taylor - polynomial formula: $T_2(x)=e+0\times(x-\frac{\pi}{2})+\frac{-e}{2}(x - \frac{\pi}{2})^2=e-\frac{e}{2}(x - \frac{\pi}{2})^2$.

Step4: Evaluate $T_2(1.5)$

Substitute $x = 1.5$ into $T_2(x)$: $T_2(1.5)=e-\frac{e}{2}(1.5-\frac{\pi}{2})^2$. $1.5-\frac{\pi}{2}\approx1.5 - 1.570796=-0.070796$. $(1.5-\frac{\pi}{2})^2\approx0.005012$. $T_2(1.5)=e-\frac{e}{2}\times0.005012\approx2.718282-\frac{2.718282}{2}\times0.005012\approx2.718282 - 0.006821=2.711461\approx2.71$.

Step5: Evaluate $f(1.5)$

$f(1.5)=e^{\sin(1.5)}$, $\sin(1.5)\approx0.997495$, $f(1.5)=e^{0.997495}\approx2.710848$.

Step6: Compute the error

$|f(1.5)-T_2(1.5)|=|2.710848 - 2.71|=0.000848\approx0.000848$.

Answer:

$T_2(1.5)=2.71$ $|f(1.5)-T_2(1.5)|=0.000848$