Download HW#1 §1.1 #10, 22, 24, 32, 38, 46, 47 In Problems 5 through 10, find

HW#1 §1.1 #10, 22, 24, 32, 38, 46, 47
In Problems 5 through 10, find all values of a such that g(a)=5.
(10) g(x)=2x2–x+4
Setting g(a)=5 we get 2a2–a+4=5 or (2a+1)(a–1)=2a2–a–1=0. Hence a=1 or
a=–1/2.
In Problems 21 through 35, find the largest domain (of real-numbers) on which the
given formula determines a (real-valued) function.
(22) f(x)=x3+5
This is a polynomial, therefore it is defined for all values of x. Hence the domain is
R = (– ∞,∞), the entire real numbers.
(24) g(t) =
( t)
2
You cannot take a square root of a negative number, therefore this function is only
defined for t ≥ 0. The domain is {t : t ≥ 0} = [0, ∞).
(32) h(z) =
1
4 − z2
If z2 > 4, that is if z<–2 or z>2, then we are trying to take a square root of a
2
negative number which is not possible. Also if z =4, that is if z=±2, then the
denominator is zero and we are trying to divide by zero. Otherwise the formula is fine.
Thus the domain is (–2,2) = {z: –2 < z < 2}.
(38) Express the volume V of a sphere as a function of its surface area S.
If the radius is r, then the volume is V=4πr3/3 and the surface area is S=4πr2.
Solving for r as a function of S gives r= S / 4π and plugging this into the formula
for V gives V(S)=S3/2/(3 4π ).
(46) A rectangular box has total surface area 600 cm2 and a square base with edge
length x centimeters. Express the volume V of the box as a function of x.
Measure all lengths in cm, areas in cm2 and volumes in cm3. Let h be the height
of the box (in centimeters). Then the box has two sides (the top and bottom) which are
squares of side length x (hence area x2) and four sides that are rectangles of width x
and height h (hence area xh). Thus the total surface area is 600=2x2+4xh. Hence h
is related to x by the formula h=150/x – x/2. Therefore the volume (in cm3) is
V(x)= x2h=x2(150/x – x/2)=150 x–x3/2.
(47) An open-topped box is to be made from a square piece of cardboard of edge
length 50 in. First four small squares, each of edge length x inches, are cut from the
corners of the cardboard. Then the four resulting flaps are turned up to form the four
sides of the box, which will thus have a square base and a depth of x inches.
Express the volume V as a function of x.
The base is a square of side length 50–2x since we lose x inches off of each
side of the cardboard. Therefore the volume is
V=x(50–2x)2.