# Prove \$ sin (x)> x – frac {x ^ 3} {3!} \$ On \$ (0, sqrt {20}) \$

I have a bit of trouble with that because my proof attempt is breaking. Down.

Evidence:
Just show that $$f (x) = sin (x) – x + frac {x ^ 3} {3!}> 0$$ sure $$I = (0, sqrt {20})$$. This is true if $$f (x)$$ strictly increases on the interval and $$f (0) geq 0$$. We can also apply this property to the first and second derivatives. We note $$f ^ {(3)} (x) = – cos (x) + 1$$. However, I can only show that $$f ^ {(3)} (x) geq 0$$ sure $$I$$ from home $$x = frac {3 pi} {2}$$ $$f ^ {(3)} (x) = 0$$