Prove that $f$ has a simple zero at $x_0$ if and only if: $$f(x) =g(x)(x-x_0),$$

where $g$ is continuous at $x_0$ and differentiable on a deleted neighborhood of $x_0$, and $g(x_0)\neq{0}$

I know that a function has a simple zero at $x_0$ if $f$ is differentiable at $x_0$ and $f(x_0) = 0$, while $f'(x_0)\neq{0}$

But I am unsure how to start on the proof. Any help is appreciated.


If $f$ has a simple zero at $x_0$, then define$$g(x)=\begin{cases}\frac{f(x)-f(x_0)}{x-x_0}&\text{ if }x\neq x_0\\f'(x_0)&\text{ otherwise.}\end{cases}$$Then $f(x)=g(x)(x-x_0)$ and $g$ is continuous at $x_0$, since$$g(x_0)=f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}g(x).$$Besides, $g(x_0)=f'(x_0)\neq0$.

On the other hand, if $f(x)=g(x)(x-x_0)$ and $g$ is continuous at $x_0$, with $g(x_0)\neq0$, then$$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}g(x)=g(x_0)\neq0.$$

  • $\begingroup$ Looks good, I'm just having trouble understanding how you showed $f(x)=g(x)(x-x_0)$ $\endgroup$ – A curious one Nov 6 '17 at 22:49
  • 1
    $\begingroup$ If $x=x_0$, then $f(x_0)=0=g(x_0)(x_0-x_0)$. Otherwise,$$g(x)(x-x_0)=\frac{f(x)-f(x_0)}{x-x_0}(x-x_0)=f(x)-f(x_0)=f(x).$$ $\endgroup$ – José Carlos Santos Nov 6 '17 at 22:52

If $f(x)=g(x)(x-x_0)$, then differentiate and trivially get the conclusion that $x_0$ is a simple zero. Conversely, let $f$ has a simple zero at $x_0$. By a desired formula $f(x)=g(x)(x-x_0)$, $g$ should be defined by $g(x)=\dfrac{f(x)}{x-x_0}.$ By the definition of a derivative $f'(x_0)$, or, as you prefer, Hospital Rule, define $g(x_0)=f'(x_0)$. Now check the details.


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