Find absolute extrema of $f(x)=\begin{cases}x^2-1,&x<1,\\\ln(x),&x\geq1\end{cases}$ 
Find absolute extrema of $$f(x)=\begin{cases}x^2-1,&x<1,\\\ln(x),&x\geq1.\end{cases}$$


I could find the critical points by setting $f'(x)=0$: $$f'(x)=\begin{cases}2x,&x<1,\\\dfrac{1}{x},&x\geq1\end{cases}=0\implies2x=0\implies x=0,$$ and $$f''(x)=\begin{cases}2,&x<1,\\-\dfrac{1}{x^2},&x\geq1\end{cases}\implies f''(0)=2>0,$$ therefore $x=0$ is a relative minimum of $f$, which has value $f(0)=-1$:

How can we prove that it is also an absolute minimum?
I tried the following:
It is not hard to see that $x^2-1\geq-1$ for $x<1$, and $\ln(x)\geq0$ for $x\geq1$. Combining these, $x=-1$ is the first of the upper bounds of $f$, so $x=-1$ is an absolute minimum.
Is it correct? If not, how would you prove it?
Thanks!!
 A: I think you have the right idea but it is phrased poorly. What you need to show is that for every $x \in \Bbb{R}$, $f(x) \geq f(0)$. So I would phrase it as follows. Let $x \in \Bbb{R}$ be arbitrary.
Case $1$: $x < 1$
In this case, 
\begin{align}
f(x) = x^2 -1 \geq -1 = f(0)
\end{align}
Case $2$: $x \geq 1$
In this case, since $\ln$ is increasing, we have that
\begin{align}
f(x) = \ln(x) \geq \ln(1) = 0 > -1 = f(0)
\end{align}
So, in every case, we have $f(x) \geq f(0)$, proving that $0$ is an absolute minimum point, and $f(0) = -1$ is the absolute minimum value.
A: Your function $f$ is piecewise defined, namely $f(x)=x^2-1$ if $x<1$ and $f(x)=\ln(x)$ if $x\ge1$. You have already found that $x=0$ is an extremal point, in particular a local minimum. 
Now you know that $\ln(x)\ge0$ for any $x\ge1$ and it is monotone crescent, hence its minimum in the interval $[1,\infty[$ is taken only in $x=1$. On the other hand, $x=1$ is not an extremal point of $f$. Hence, if another local minimum existed, it would be also a local minimum for $x^2-1$. 
You know that $x^2-1$ is convex and the minimum is taken only in $x=0$. Hence you can conclude that $x=0$ is the absolute minimum of $f$. 
