In this this review we turn our attention to the square root function, the function defined by the equation. f(x) = x. (5.1)


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1 Section 5.2 The Square Root The Square Root In this this review we turn our attention to the square root function, the function defined b the equation f() =. (5.1) We can determine the domain and range of the square root function b projecting all points on the graph onto the  and aes, as shown in Figures 1(a) and (b), respectivel. f f (a) Domain = [0, ) (b) Range = [0, ) Figure 1. Project onto the aes to find the domain and range. Translations If we shift the graph of = right and left, or up and down, the domain and/or range are affected. Eample 1 Determine the domain and range of f() = 2. If we replace with 2, the basic equation = becomes = 2. From our review work with transformations, we know that this will shift the graph two units to the right, as shown in Figures 2(a) and (b). To find the domain, we project each point on the graph of f onto the ais, as shown in Figure 2(a). Note that all points to the right of or including 2 are shaded on the ais. Consequentl, the domain of f is Domain = [2, ) = { : 2}. As there has been no shift in the vertical direction, the range remains the same. To find the range, we project each point on the graph onto the ais, as shown in Figure 2(b). Note that all points at and above zero are shaded on the ais. Thus, the range of f is
2 2 Chapter 5 f f (a) Domain = [2, ) (b) Range = [0, ) Figure 2. To draw the graph of f() = 2, shift the graph of = two units to the right. Range = [0, ) = { : 0}. We can find the domain of this function algebraicall b eamining its defining equation f() = 2. We understand that we cannot take the square root of a negative number. Therefore, the epression under the radical must be nonnegative (positive or zero). That is, 2 0. Solving this inequalit for, 2. Thus, the domain of f is Domain = [2, ), which matches the graphical solution above. The Pthagorean Theorem Our review now focus s on a ver famous and practical theorem
3 Section 5.2 The Square Root 3 Pthagorean Theorem. Let c represent the length of the hpotenuse, the side of a right triangle directl opposite the right angle (a right angle measures 90 ) of the triangle. The remaining sides of the right triangle are called the legs of the right triangle, whose lengths are designated b the letters a and b. c b a The relationship involving the legs and hpotenuse of the right triangle, given b a 2 + b 2 = c 2, (5.2) is called the Pthagorean Theorem. Note that the Pthagorean Theorem can onl be applied to right triangles. Let s look at a simple application of the Pthagorean Theorem (5.2). Eample 2 Given that the length of one leg of a right triangle is 4 centimeters and the hpotenuse has length 8 centimeters, find the length of the second leg. Let s begin b sketching and labeling a right triangle with the given information. We will let represent the length of the missing leg. 8 cm 4 cm Figure 3. bit easier. A sketch makes things a Here is an important piece of advice.
4 4 Chapter 5 Tip 1 The hpotenuse is the longest side of the right triangle. It is located directl opposite the right angle of the triangle. Most importantl, it is the quantit that is isolated b itself in the Pthagorean Theorem. a 2 + b 2 = c 2 Alwas isolate the quantit representing the hpotenuse on one side of the equation. The legs go on the other side of the equation. So, taking the tip to heart, and noting the lengths of the legs and hpotenuse in Figure 3, we write = 8 2. Square, then isolate on one side of the equation = 64 2 = 48 Normall, we would take plus or minus the square root in solving this equation, but represents the length of a leg, which must be a positive number. Hence, we take just the positive square root of 48. = 48 Of course, place our answer in simple radical form. = 16 3 = 4 3 If need be, ou can use our graphing calculator to approimate this length. To the nearest hundredth of a centimeter, 6.93 centimeters. Applications of the Pthagorean Theorem Let s look at a few eamples of the Pthagorean Theorem in action. The practical applications of the Pthagorean Theorem are numerous. Eample 3 A painter leans a 20 foot ladder against the wall of a house. The base of the ladder is on level ground 5 feet from the wall of the house. How high up the wall of the house will the ladder reach? Consider the triangle in Figure 4. The hpotenuse of the triangle represents the ladder and has length 20 feet. The base of the triangle represents the distance of the base of the ladder from the wall of the house and is 5 feet in length. The vertical leg of the triangle is the distance the ladder reaches up the wall and the quantit we wish to determine.
5 Section 5.2 The Square Root 5 20 h Appling the Pthagorean Theorem, 5 Figure 4. A ladder leans against the wall of a house h 2 = Again, note that the square of the length of the hpotenuse is the quantit that is isolated on one side of the equation. Net, square, then isolate the term containing h on one side of the equation b subtracting 25 from both sides of the resulting equation h 2 = 400 h 2 = 375
6 6 Chapter 5 We need onl etract the positive square root. h = 375 We could place the solution in simple form, that is, h = 5 15, but the nature of the problem warrants a decimal approimation. Using a calculator and rounding to the nearest tenth of a foot, h Thus, the ladder reaches about 19.4 feet up the wall. The Distance Formula We often need to calculate the distance between two points P and Q in the plane. Indeed, this is such a frequentl recurring need, we d like to develop a formula that will quickl calculate the distance between the given points P and Q. Such a formula is the goal of this last section. Let P ( 1, 1 ) and Q( 2, 2 ) be two arbitrar points in the plane, as shown in Figure 5(a) and let d represent the distance between the two points. Q( 2, 2 ) Q( 2, 2 ) d d 2 1 P ( 1, 1 ) P ( 1, 1 ) 2 1 R( 2, 1 ) (a) Figure 5. Finding the distance between the points P and Q. To find the distance d, first draw the right triangle P QR, with legs parallel to the aes, as shown in Figure 5(b). Net, we need to find the lengths of the legs of the right triangle P QR. The distance between P and R is found b subtracting the coordinate of P from the coordinate of R and taking the absolute value of the result. That is, the distance between P and R is 2 1. The distance between R and Q is found b subtracting the coordinate of R from the coordinate of Q and taking the absolute value of the result. That is, the distance between R and Q is 2 1. (b)
7 Section 5.2 The Square Root 7 We can now use the Pthagorean Theorem to calculate d. Thus, However, for an real number a, d 2 = ( 2 1 ) 2 + ( 2 1 ) 2. ( a ) 2 = a a = a 2 = a 2, because a 2 is nonnegative. Hence, ( 2 1 ) 2 = ( 2 1 ) 2 and ( 2 1 ) 2 = ( 2 1 ) 2 and we can write d 2 = ( 2 1 ) 2 + ( 2 1 ) 2. Taking the positive square root leads to the Distance Formula. The Distance Formula. Let P ( 1, 1 ) and Q( 2, 2 ) be two arbitrar points in the plane. The distance d between the points P and Q is given b the formula d = ( 2 1 ) 2 + ( 2 1 ) 2. (5.3) The direction of subtraction is unimportant. Because ou square the result of the subtraction, ou get the same response regardless of the direction of subtraction (e.g. (5 2) 2 = (2 5) 2 ). Thus, it doesn t matter which point ou designate as the point P, nor does it matter which point ou designate as the point Q. Simpl subtract  coordinates and square, subtract coordinates and square, add, then take the square root. Let s look at an eample. Eample 4 Find the distance between the points P ( 4, 2) and Q(4, 4). It helps the intuition if we draw a picture, as we have in Figure 6. One can now take a compass and open it to the distance between points P and Q. Then ou can place our compass on the horizontal ais (or an horizontal gridline) to estimate the distance between the points P and Q. We did that on our graph paper and estimate the distance d. Let s now use the distance formula to obtain an eact value for the distance d. With ( 1, 1 ) = P ( 4, 2) and ( 2, 2 ) = Q(4, 4), d = ( 2 1 ) 2 + ( 2 1 ) 2 = (4 ( 4)) 2 + (4 ( 2)) 2 = = = 0 =.
8 8 Chapter 5 It s not often that our eact result agrees with our approimation, so never worr if ou re off b just a little bit. Q(4, 4) d P ( 4, 2) Figure 6. Gauging the distance between P ( 4, 2) and Q(4, 4).
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