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I apologize for the specificity of the my question, but I'm concerned that I'm having trouble grasping an important concept.

I'm puzzled by the answer provided for exercise 1.(v) in chapter 7 of Spivak's Calculus (4E, p.129):

For $a>-1$ and $f(x) =\begin{cases} x^{2}, & x ≤ a \\ a+2, & x>a \end{cases},\qquad x\in(-a-1,a+1),$

where where does $f(x)$ take on its maximum and minimum?

I get $\begin{array}{cc} Range & Max & Min\\ -1 but the answer key has $a^{2}$ as a minimum only for $-\frac{1}{2}.

What am I missing?

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    @Arturo: Exactly that. Thanks for setting me straight!2011-09-11

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I have the 2nd Spanish edition (Editorial Reverté, S.A.), translated from the second English edition. The problem is the same, but did not include the condition $a\gt -1$ until the answer key. But the answer key there reads (translated):

It is bounded above and below. It is understood that $a\gt -1$ (so that $-a-1\lt a+1$). If $-1\lt a\leq -\frac{1}{2}$, then $a\lt -a-1$, so $f(x)=a+2$ for all $x\in (-a-1,a+1)$, so $a+2$ is the maximum and the minimum. If $-\frac{1}{2}\lt a\leq 0$, then $f$ has minimum $a^2$, and if $a\geq 0$, then it has minimum $0$. Since $a+2\gt (a+1)^2$ only for $\frac{-1-\sqrt{5}}{2}\lt a \lt \frac {1+\sqrt{5}}{2}$, when $a\geq -\frac{1}{2}$ the function $f$ has a maximum only for $a\leq \frac{1+\sqrt{5}}{2}$ (when this maximum is $a+2$).

So it looks like your answer matches exactly with this one.

Added. Oh, I see; the problem is what happens when $a=-\frac{1}{2}$.

If $a=-\frac{1}{2}$, then the function is $f(x) = \left\{\begin{array}{ll}x^2 & \text{if }x\leq -\frac{1}{2}\\ \frac{3}{2} &\text{if }x\gt -\frac{1}{2} \end{array}\right.\qquad x\in\left(-\left(-\frac{1}{2}\right)-1,-\frac{1}{2}+1\right).$ What you seem to be missing is that since the domain is $(-\frac{1}{2},\frac{1}{2})$, the first case never occurs, so you are always in the second case.

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    Ah — that's it: if $a=-\frac{1}{2}$, $a$ isn't in the domain of $f(x)$!2011-09-11