$|f(x)-f(1)|<k|x-1|$

Question

Given the function :

$f(x)=x^2+x|x-1|-1$ such that $x$ is a real number .

Show that there is a $k\in \mathbb{R}$ for all $x\in \mathbb{R}$ such that:

$|x-1|<1 \implies |f(x)-f(1)|<k|x-1|$

I tried cases of $x$ :

If $x>1$ then $f(x)=(2x+1)(x-1)$

If $x<1$ then $f(x)=x-1$

And so the $k=2x+1$ ?

And then conclude that for all positive $\varepsilon$ there is a positive $\alpha$ for all real $x$ Such that:

$|x-1|\lt1 \implies |f(x)-f(1)|\lt\varepsilon$

Answer 1

If $x \in \mathbb{R}$, then $$ |f(x) - f(1)| = |x^{2}+x|x-1|-1| \leq |x-1|(|x| + |x+1|); $$ if $|x-1| < 1$, then $0 < x < 2$, so $$ |x-1|(|x| + |x+1|) < 5|x-1|; $$ take $k := 5$.