Mock Midterm 1BNote: The problems on this mock midterm were not necessarily selected to allow them tobe easy to work without a calculator. The problems on the real midterm will not require acalculator.(1) As advances in technology result in the production of increasingly powerful, compactcalculators, the price of calculators currently on the market drops. Suppose that xmonths from now, the price of a certain model will be P dollars per unit, whereP (x) = 40 +30x + 1.(a) What will the price be 5 months from now?P (5) = 40 +305 + 1= 40 +306= 40 + 5 = 45.(b) By how much will the price drop during the fifth month?To find how much the price drops during the 5th month, we need to computethe prices after 4 months and after 5 months and find the difference. We alreadyknow that P (5) = 45. We computeP (4) = 40 +304 + 1= 40 +305= 40 + 6and so during the 5th month the price drops $46 − $45 = $1.(c) When will the price be $43?To find when the price will be $43, we set P (x) = 43 and solve for x. We have43 = 40 +30x + 1⇒ 3 =30x + 1⇒ 3x + 3 = 30 ⇒ 3x = 27 ⇒ x = 9.And so the price will be $43 in 9 months.(2) Find equations for these lines:(a) Passes through the points (−1, 3) and (4, 1).We begin by computingm =3 − 1−1 − 4= 2−5.This, together with the slope-point form of a line, allows us to computey − y0= m(x − x0) ⇒ y − 3 =−25(x − −1) ⇒ y =−25x +135.(b) x intercept (3, 0) and y intercept (0, −2/3).We again begin by computingm =0 − −2/33 − 0⇒23 · 3⇒29.Since we know the y intercept, we can use point-intercept form to findy = mx + b ⇒ y = (2/9)x − 2/3.1(c) Contains (−1, 3) and is perpendicular to 5x − 3y = 7.We need a line which is perpendicular to 5x − 3y = 7, so we need to first findthe slope m of the given line. The slope of our desired line will then be given by−1/m. Solving the equation for y, we have5x − 3y = 7 ⇒ −3y = 7 − 5x ⇒ y = (5/3)x − 7/3and so m = 5/3 and −1/m = −3/5. We can then use slope-point form to gety−y0= m(x−x0) ⇒ y−3 = (−3/5)(x−−1) ⇒ y = (−3/5)x−(3/5)+3 ⇒ y = (−3/5)x+12/5.(3) (a) Define continuity.A function f is continuous at x = c if all three of these conditions are satisfied:(i) f(c) is defined,(ii) limx→cf(x) exists, and(iii) limx→cf(x) = f(c).(b) List all values for whichf(x) =(x3+ 2x − 33 x ≤ 3x2−6x+9x−3x > 3is not continuous.We need to check the points where each of the individual pieces of the functionare not defined, and also the point at which we switch from one definition of thefunction to the other. Since our first function is a polynomial, it is continuouseverywhere. Our second function is not defined at x = 3, which is also, coinci-dentally, the point at which our function changes definitions. To check whetheror not our function is continuous at x = 3, we need to see whether or not it hasa limit at x = 3. We havelimx→3−f(x) = limx→3−x3+ 2x − 33= (3)3+ 2 · 3 − 33= 27 + 6 − 33= 0= f(3)andlimx→3+f(x) = limx→3+x2− 6x + 9x − 3= limx→3+(x − 3)(x − 3)x − 3= limx→3+x − 3= 3 − 3= 0So limx→3f(x) = 0 = f(3), and so our function has no points of discontinuity.2(4) Find each of these limits. If the limit is infinite, indicate whether it is +∞ of −∞.(a) limx→−1(x2+ 2x − 3)/(x − 1)limx→−1x2+ 2x − 3x − 1=(−1)2+ 2(−1) − 3−1 − 1=1 − 2 − 3−2= 2(b) limx→1(x2+ 2x − 3)/(x − 1)limx→1x2+ 2x − 3x − 1= limx→1(x + 3)(x − 1)x − 1= limx→1x + 3 = 4.(c) limx→1(x2− x − 1)/(x − 2)limx→1x2− x − 1x − 2=12− 1 − 11 − 2=−1−1= 1(d) limx→+∞(2x3+ 3x − 5)/(−x2+ 2x + 7)limx→+∞2x3+ 3x − 5−x2+ 2x + 7= limx→+∞2x3+ 3x − 5−x2+ 2x + 7·1x21x2= limx→+∞2x +3x−5x2−1 +2x+7x2= −∞(5) Find the derivatives of the following functions.(a) f(x) = x3+ 3x2− 7f0(x) = 3x3−1+ 3 · 2x2−1− 0= 3x2+ 6x(b) g(x) = (x2+ 2)(3x −2x)g0(x) = (x2+ 2)0(3x − 2x−1) + (x2+ 2)(3x − 2x−1)0= (2x)(3x − 2x−1) + (x2+ 2)(3 + 2x−2)= 6x2− 4 + 3x2+ 2 + 6 + 4x−2= 4x−2+ 9x2+ 43(c) h(x) = (3x + 1)/(x2+ 2x)h0(x) =(3x + 1)0(x2+ 2x) − (3x + 1)(x2+ 2x)0[(x2+ 2x)]2=3(x2+ 2x) − (3x + 1)(2x + 2)[(x2+ 2x)]2=3x2+ 6x − [6x2+ 6x + 2x + 2][(x2+ 2x)]2=−3x2− 2x − 2[(x2+ 2x)]2(6) Khalil is trying to decide between two competing property tax propositions. WithProposition A, he will pay $100 plus 8% of the assessed value of his home, whileProposition B requires a payment of $1900 plus 2% of the assessed value. AssumingKhalil’s only consideration is to minimize his tax payment, develop a criterion basedon the assessed value V of his home for deciding between the propositions.TA(x) = 100 + .08xandTB(x) = 1900 + .02x.Clearly TA(x) is the better option when the assessed value of the house is $0. Weneed to find at what point TB(x) is the better option, and in order to do this we needto find the point of intersection of the two taxing lines. We haveTA(x) = TB(x)100 + .08x = 1900 + .02x.06x = 1800x = 1800/.06 = 30000.So for an assessed value of $30000, the two tax propositions would be equal, but forvalues of more than $30000 proposition B would be the better option and for valuesless than $30000 proposition A would be the better