Question

How would I find the fourth-order Taylor polynomial for the function?

Answer 1

Answer:

7889/6114

Step-by-step explanation:

$e^x \approx 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!} \\ \\ =1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24} \\ \\ \implies e^{3x} \approx 1+3x+\frac{(3x)^2}{2}+\frac{(3x)^3}{6}+\frac{(3x)^4}{24} \\ \\ =1+3x+\frac{9}{2}x^2+\frac{9}{2}x^3+\frac{27}{8}x^4 \\ \\ \implies e^{1/4} \approx 1+3(1/12)+\frac{9}{2}(1/12)^2+\frac{9}{2}(1/12)^3+\frac{27}{8}(1/12)^4 \\ \\ =\frac{7889}{6144}$