## Calculus (3rd Edition)

$$4.3258667 \times 10^{-6}$$
Since \begin{array}{ll} {f(x)=x^{-\frac{1}{2}},} & {f(4)=\frac{1}{2}} \\ {f^{\prime}(x)=-\frac{1}{2} x^{-\frac{3}{2}},} & {f^{\prime}(4)=-\frac{1}{16}} \\ {f^{\prime \prime}(x)=\frac{3}{4} x^{-\frac{5}{2}},} & {f^{\prime \prime}(4)=\frac{3}{128}} \\ {f^{\prime \prime \prime}(x)=-\frac{15}{8} x^{-\frac{7}{2}},} & {f^{\prime \prime \prime}(4)=-\frac{15}{1024}} \end{array} Then $$T_{3}(x)=\frac{1}{2}-\frac{1}{16}(x-4)+\frac{3}{256}(x-4)^{2}-\frac{5}{2048}(x-4)^{3}$$ Since $$\left|f(x)-T_{n}(x)\right| \leq K \frac{|x-a|^{n+1}}{(n+1) !}$$ Then \begin{align*} \left|f(4.3)-T_{3}(4.3)\right| &\leq \frac{105}{8192} \frac{|4.3-4|^{3+1}}{(3+1) !} \\ \left|f(4.3)-T_{3}(4.3)\right| &\leq \frac{105}{8192} \frac{(0.3)^{4}}{(4) !}\\ &=\frac{105}{8192} \frac{(0.3)^{4}}{(24)} \\ &=\frac{105}{8192} \frac{(0.3)^{4}}{(24)} \\ &\approx\left[4.3258667 \times 10^{-6}\right] \end{align*} To check \begin{aligned} \left|f(4.3)-T_{3}(4.3)\right| & \approx|0.4822428222-0.4822387695| \\ & \approx 4.0527 \times 10^{-6}<4.3258667 \times 10^{-6} \end{aligned}