Question

f(x)=X over x^3-2x^2+5x why will this have no zeros​

Answer 1

If you evaluate directly this function at x=0, you'll see that you have a zero denominator.

Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

$\dfrac{x}{x^3-2x^2+5x}=0\iff x=0$

So, this function can't have zeroes, because the only point that would annihilate the numerator would annihilate the denominator as well.

Moreover, we have

$\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}$

So, we can't even extend with continuity this function in such a way that $f(0)=0$