There is already a very helpful post about part (v) of this question on this site here . I thought that the answer to problem (vi) would be similar with minor differences, but somehow it is not. The problem reads as follows:
1. For each of the following functions, decide which are bounded above or below on the indicated interval, and which take on their maximum or minimum value. (Notice that $f$ might have these properties even if $f$ is not continuous, and even if the interval is not a closed interval.)
- (vi) $f(x)=\begin{cases} x^2,& x<a\\ a+2,& x\geq a \end{cases} \mbox{ on } [-a-1, a+1].$
Spivak's answer:
(vi) Bounded above and below. As in part (v), it is assumed that $a> -1$. If $a \leq -\frac 12$, then $f$ has the minimum and maximum value $\frac 32$. If $ a \geq 0$, then $f$ has the minimum value $0$, and the maximum value $max(a^2, a+2)$. If $-\frac 12 <a<0$, then $f$ has the maximum value $\frac 32$ and no minimum value.
My objections to his solution:
- If $a=-1$, then $[-a-1, a+1]$ contains $0$. So for $a=-1$, $f(0)=a+2=-1+2=1$.
- If $a\leq -\frac 12$, for example let $a=-\frac 23$, then $[-a-1, a+1]=[-\frac 13, \frac 13]$ and $f(x)=a+2= -\frac 23 + 2= \frac 43$ for all $x$ of that interval. In this case $f$ has the minimum and maximum value $\frac 43$ and not $\frac 32$. (Maybe he meant that the "maximum" minimum and maximum value for $a\leq -\frac 12$, which appear for $a=-\frac 12$, is $\frac 32$, but I am not sure…)
- If $a\geq 0$, for example let $a=3$, then $[-a-1, a+1]=[-4, 4]$ and $f(-4)=16>a^2=9>a+2=5$. $f$ has the maximum value $(-a-1)^2$ for this $a$, so I suspect on this issue $f$ behaves similar to the function of part (v).
- If $-\frac 12<a<0$, for example let $a=-\frac{4}{10}$, then $[-a-1, a+1]=[-\frac{6}{10}, \frac{6}{10}]$ and $f(0)=a+2=-\frac{4}{10}+2=\frac{16}{10}>\frac{3}{2}$. In this case $f$ has the maximum value $\frac{16}{10}$ and not $\frac{3}{2}$.
Am I missing something or is there a mistake in every sentence of Spivak's answer?
Best Answer
Spivak's answer to the previous step (v) says
This makes sense especially for the open interval $(-a-1,a+1)$ in that part of the problem.
However, for this part of the problem (which uses a closed interval) we can instead allow $a = -1$, in which case your solution would be correct.
Spivak's repeated use of $3/2$ instead of $a+2$ appears to be a typo. I think you're right there as well.
His answer misses another thing probably worth mentioning: if $a = 0$, the interval becomes $[-1, 1]$. Here, $f$ approaches the minimum value of $0$ from the left, but never takes on that value.