Question

If $f\left(x\right)=\frac{x^3+3x+5}{x^3-3x+2}$. Discuss the continuity of f(x) on [0,5].

Answer 1

$f\left(x\right)=\frac{x^3+3x+5}{x^3-3x+2}$

$f\left(0\right)=\frac{0+0+5}{0-0+2}=\frac{5}{2}$

$f\left(1\right)=\frac{1+3+5}{1-3+2}=\frac{9}{0}=\infty$

$f\left(2\right)=\frac{8+6+5}{8-6+2}=\frac{19}{4}$

$f\left(3\right)=\frac{27+9+5}{27-9+2}=\frac{41}{20}$

$f\left(4\right)=\frac{64+12+5}{64-12+2}=\frac{81}{74}$

$f\left(5\right)=\frac{125+15+5}{125-15+2}=\frac{145}{112}$

$\therefore f\left(x\right)$ is continuous at $x=0,2,3,4,5$.