Solutions

i) $f(x)=x^2$ on $(-1,1)$ is bounded above and below. It does not take on its max but does take on its min.

ii) $f(x)=x^3$ on $(-1,1)$ is bounded but neither takes on a max nor a min.

iii) $f(x)=x^2$ on $\mathbb{R}$ is bounded below but not above. It takes on a min but not a max.

iv) $f(x)=x^2$ on $[0,\infty)$ has the same analysis as in iii).

v) $f$ on $(-a-1,a+1)$ is bounded for all $a>-1$. For $a\in(-1,-\frac12)$, we have $a<-a-1$; hence, $f(x)=a+2$, so it takes on both its max and its min. For $a\in[-\frac12,0)$, $\min f(x)=a^2$. For $a\in[0,\infty)$, $\min f(x)=0$. For $a\in[-\frac12,\frac{-1+\sqrt{5}}2]$, $\max f(x)=a+2$. For $a\in(\frac{-1+\sqrt{5}}2,\infty)$, there is no max.

vi) $f$ on $[-a-1,a+1]$ is bounded for all $a$. For $a\in(-1,-\frac12]$, $f(x)=a+2$, so it takes on both its min and max. For $a\in(-\frac12,0]$, there is no minimum. For $a\in(0,\infty]$, $\min f(x)=0$. For $a\in(-\frac12,\frac{-1+\sqrt{5}}2]$, $\max f(x)=a+2$. For $a\in(\frac{-1+\sqrt{5}}2,\infty)$, $\max f(x)=(a+1)^2$.

vii) $f$ takes a min value of $0$ and a max value of $1$. Hence, it is also bounded.

viii) $f$ takes on a max value of $1$ but does not have a min value. It is bounded, however.

ix) $\min f=-1$ and $\max f=1$. Hence, $f$ is bounded.

x) For all $a>0$, $\max f=a$ and $\min f=0$. Hence, $f$ is bounded.

xi) Assuming $f$ is continuous, it has a min and a max by Theorems 3 and 7. Hence, it is bounded.

xii) Bounded; has a min and a max.