Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$\begin{array}{cc} Range & Max & Min\\ -1<a<-\frac{1}{2} & a+2 & a+2 \\ -\frac{1}{2}≤a<0 & a+2 & a^{2} \\ 0≤a≤\frac{\sqrt{5}-1}{2} & a+2 & 0 \\ \frac{\sqrt{5}-1}{2}<a & - & 0 \\ \end{array}$$ but the answer key has $a^{2}$ as a minimum only for $-\frac{1}{2}<a\lt 0$.

What am I missing?

share|improve this question
1  
When you say $f(x)=x^2$ for $x \le a$, do you mean $f(x)=a^2$? Otherwise very negative $x$ will produce a large $x^2$. –  Henry Sep 11 '11 at 21:45
1  
@Henry: I only have the 2nd Edition, and in Spanish, but that's what it says in that edition. Note that the domain is restricted to $(-a-1,a+1)$, so "very negative $x$" will not be in the domain. –  Arturo Magidin Sep 11 '11 at 21:52
    
@Henry: as written. Though maybe I should clarify: what the answer key does is shift the case of $a=-\frac{1}{2}$ from the second to the first line my solution above. –  raxacoricofallapatorius Sep 11 '11 at 21:54
1  
@raxacoricofallapatorius: I see! The problem is that when $a=-\frac{1}{2}$, the domain is $(-\frac{1}{2},\frac{1}{2})$, so you are never in the case $x\leq a$; the function is constant $a+2$ in that case. –  Arturo Magidin Sep 11 '11 at 22:43
    
@Arturo: Exactly that. Thanks for setting me straight! –  raxacoricofallapatorius Sep 11 '11 at 23:31
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
Ah — that's it: if $a=-\frac{1}{2}$, $a$ isn't in the domain of $f(x)$! –  raxacoricofallapatorius Sep 11 '11 at 22:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.