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.

`(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) (x)/(x^3-2x^2+5x) = \lim_(x\to 0) (x)/(x(x^2-2x+5)) = \lim_(x\to 0) (1)/(x^2-2x+5) = (1)/(5)`

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