7. Show that the expression $x+{\frac {1}{x}}$ cannot take on any value strictly between 2 and -2.

$f(x)=x+{\frac {1}{x}}$ $f'(x)=1-{\frac {1}{x^{2}}}$ $1-{\frac {1}{x^{2}}}=0\implies x=\pm 1$ $f''(x)={\frac {2}{x^{3}}}$ $f''(-1)=-2$
Since $f''(-1)$ is negative, $x=-1$ corresponds to a relative maximum. $f(-1)=-2$ $\lim \limits _{x\to -\infty }f(x)=-\infty$
For $x<-1$ , $f'(x)$ is positive, which means that the function is increasing. Coming from very negative $x$-values, $f$ increases from a very negative value to reach a relative maximum of $-2$ at $x=-1$ .
For $-1<x<1$ , $f'(x)$ is negative, which means that the function is decreasing. $\lim _{x\to 0^{-}}f(x)=-\infty$ $\lim _{x\to 0^{+}}f(x)=\infty$ $f''(1)=2$
Since $f''(1)$ is positive, $x=1$ corresponds to a relative minimum. $f(1)=2$
Between $[-1,0)$ the function decreases from $-2$ to $-\infty$ , then jumps to $+\infty$ and decreases until it reaches a relative minimum of $2$ at $x=1$ .
For $x>1$ , $f'(x)$ is positive, so the function increases from a minimum of $2$ .
The above analysis shows that there is a gap in the function's range between $-2$ and $2$ .

22. You peer around a corner. A velociraptor 64 meters away spots you. You run away at a speed of 6 meters per second. The raptor chases, running towards the corner you just left at a speed of $4t$ meters per second (time $t$ measured in seconds after spotting). After you have run 4 seconds the raptor is 32 meters from the corner. At this time, how fast is death approaching your soon to be mangled flesh? That is, what is the rate of change in the distance between you and the raptor?

$10{\tfrac {m}{s}}$

23. Two bicycles leave an intersection at the same time. One heads north going $12{\rm {mph}}$ and the other heads east going $5{\rm {mph}}$ . How fast are the bikes getting away from each other after one hour?

$13{\rm {mph}}$

24. You're making a can of volume $200m^{3}$ with a gold side and silver top/bottom. Say gold costs 10 dollars per m$^{2}$ and silver costs 1 dollar per $m^{2}$ . What's the minimum cost of such a can?