Taylor Series

 $P_4\left(x\right)=x-\frac{1}{6}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7$  

 $P_4\left(x\right)=-x^2+\frac{1}{6}x^4$  

 $P_4\left(x\right)=x^2-\frac{1}{6}x^4$  

 $P_4\left(x\right)=2x^2-4x^4$  

 $T_4\left(x\right)=1+5x+10x^2+10x^3+5x^4$  provides a good approximation for f(x) only for values of x that are close to x = 0.

 $T_4\left(x\right)=1+5x+10x^2+10x^3+5x^4$  provides a good approximation for f(x) for all real values of x.

 $T_4\left(x\right)=1+5x+20x^2+60x^3+120x^4$  provides a good approximation for f(x) only for values of x that are close to x = 0.

 $T_4\left(x\right)=1+5x+20x^2+60x^3+120x^4$   provides a good approximation for f(x) for all real values of x.

 $x-3-\frac{\left(x-3\right)^2}{2}-\frac{\left(x-3\right)^3}{3}$  

 $x-3-\frac{\left(x-3\right)^2}{2}+\frac{\left(x-3\right)^3}{3}$  

 $x+3+\frac{\left(x+3\right)^2}{2}+\frac{\left(x+3\right)^3}{3}$  

 $x+3-\frac{\left(x+3\right)^2}{2}+\frac{\left(x+3\right)^3}{3}$  

 $-\frac{1}{144}$  

 $-\frac{1}{72}$  

 $-\frac{1}{9}$  

 $\frac{1}{144}$  

 $9e^3$  

 $\frac{e^3}{9!}$  

 $\frac{e^9}{3!}$  

 $\frac{e^3}{9}$  

 $6-\frac{x}{5}+\frac{x^2}{7}+\frac{x^3}{6};\ \ f\left(0.5\right)\approx5.957$  

 $6-5x+7x^2+6x^3;\ \ f\left(0.5\right)\approx6.000$  

 $6x-5x^2+7x^3+6x^4;\ \ f\left(0.5\right)\approx3.000$  

 $6+7x-5x^2+6x^3;\ \ \ f\left(0.5\right)\approx9.000$