Find the minimum value of $f(x) = \max\{x^2-4, x , - 1\}$

I am able to do this question by using graph. But if there is any other method please tell me

  • $\begingroup$ I tried to improve the readability by adding $\LaTeX$ code, please let me know if I misinterpreted the function. $\endgroup$ – Asaf Karagila Jun 16 '12 at 6:41
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    $\begingroup$ The graph is the heart of the argument. One can then translate the information one sees into inequalities, and give formal proofs if desired. $\endgroup$ – André Nicolas Jun 16 '12 at 6:56

You can reason as follows. The $-1$ in $\max\{x^2-4,x,-1\}$ guarantees that $f(x)$ is always at least $-1$, so the minimum value of $f(x)$ must be at least $-1$. Is there at least one point $x$ such that $f(x)=-1$? At such a point we would have $x^2-4\le-1$ and $x\le-1$. The question is whether we can have both of these at the same time: is there a point $x\le-1$ such that $x^2-4\le-1$?

The inequality $x^2-4\le-1$ is equivalent to $x^2\le3$, or $-\sqrt3\le x\le\sqrt 3$. Since $-\sqrt3<-1$, we have a whole interval of $x$’s from which to choose: any $x\in[-\sqrt3,-1]$ will have the property that $x^2-4\le-1$ and $x\le -1$ and therefore will satisfy $f(x)=-1$.

Thus, $f(x)\ge-1$ for all $x\in\Bbb R$, and there is at least one $x\in\Bbb R$ such that $f(x)=-1$, so $-1$ must be the minimum value of $f(x)$.


Notice that $f(x) \geq -1$, $\forall x$, and $f(-1) = -1$, so $\min_{x \in \mathbb{R}} f(x) = -1$.

To find all $x$ such that $f(x) = -1$, we must characterize the set $M = \{ x | x^2-4\leq -1, \, x \leq -1 \}$. A moment's thought will show that $M=[-\sqrt{3},-1]$.


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