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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.