2 quick Q's: solve -3<1/x<=1 & find fcn for Norman window

[SushiUnlimited]

I'm having trouble on my calculus homework. The two question are:
Solve the inequality in terms of intervals: -3<1/x<=1

and

A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 18 ft, express the area A of the window as a function of the width x of the window.

With this picture: http://www.webassign.net/scalcet7/1-1-062.gif

I don't need a full answer but some hints and a push in the right direction would be nice. My textbook has a answer key that comes with it but it only works out every other problem and these question's textbook counterpart were even numbered.

 

[daon2]

Let \(\displaystyle y\) be the height of the rectangular part.

The perimeter of the window is:

\(\displaystyle 18 = \underbrace{\frac{1}{2}(2\pi\frac{x}{2})}_{\text{Semi-circle}} + \underbrace{x + 2y}_{\text{Rectangular piece}}\)

The area is:

\(\displaystyle A = \underbrace{\frac{1}{2}\pi \left(\frac{x}{2}\right)^2}_{\text{Semi-circle}} + \underbrace{xy}_{\text{Rectangle}}\)

Use the first equation to turn the second into a function of \(\displaystyle x\).

 

[soroban]

Hello, SushiUnlimited!

\(\displaystyle \text{Solve in terms of intervals: }\:\text{-}3\,<\,\dfrac{1}{x}\,\le\,1\)

The equalities occur when: .\(\displaystyle x \,=\,\text{-}\frac{1}{3}\,\text{ and }\,x = 1\)

The number line is divided into three intervals.

. . \(\displaystyle \begin{array}{ccccc} ---&\circ&---&\bullet & --- \\ & \text{-}\frac{1}{3} && 1 \end{array}\)


Test a value of \(\displaystyle x\) in each interval.


. . \(\displaystyle x = \text{-}2\!:\;\text{-}3 \,<\,\frac{1}{\text{-}2} \,\le\,1 \quad\Rightarrow\quad \text{-}3 \,<\,\text{-}\frac{1}{2}\,\le 1 \;\text{ . . . true} \)

. . \(\displaystyle x = \frac{1}{2}\!:\;\text{-}3 \,<\,\frac{1}{\frac{1}{2}} \,\le\,1 \quad\Rightarrow\quad \text{-}3 \,<\,2 \,\le\,1\;\text{ . . . false}\)

. . \(\displaystyle x = 2\!:\;\text{-}3 \,<\,\frac{1}{2}\,\le\,1 \;\text{ . . . true}\)


Therefore: .\(\displaystyle (\text{-}\infty,\text{-}\frac{1}{3}) \,\cup\,[1,\infty)\)