Mock Midterm 1ANote: 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) (a) Give the definition of the derivative.The derivative of the function f(x) with respect to x is the function f0(x) givenbyf0(x) = limh→0f(x + h) − f(x)h.(b) Use the definition to compute the derivatives of(i) f(x) = 1/x2(Hint: You’ll need to find a common denominator for thetop.)f0(x) = limh→0f(x + h) − f(x)h= limh→01(x+h)2−1x2h= limh→01(x+h)2x2x2−1x2(x+h)2(x+h)2h= limh→0x2− x2− 2xh − h2hx2(x + h)2= limh→0x2−x2− 2xh − h2hx2(x + h)2= limh→0−2xh − h2hx2(x + h)2= limh→0−2x − hx2(x + h)2=−2x − 0x2(x + 0)2=−2xx4=−2x3Observe that this agrees with the derivative computed using the powerrule.1(ii) g(x) =√x (Hint: Multiply by the conjugate√x + h +√x.)f0(x) = limh→0f(x + h) − f(x)h= limh→0√x + h −√xh= limh→0√x + h −√xh·√x + h +√x√x + h +√x= limh→0x + h − xh(√x + h +√x)(Work out the details)= limh→0x + h −xh(√x + h +√x)= limh→0hh(√x + h +√x)=1√x + 0 +√x=12√xNote: Hints will not be provided on the actual midterm.(2) Compute the derivatives of the following functions using any method. You need notsimplify the results.(a) f(x) = 1/3√xf0(x) = (x−1/3)0= (−1/3)x−1/3−1= (−1/3)x−4/3(b) g(x) = (x2+ 3)(1/x − x3)g0(x) = (x2+ 3)0(x−1− x3) + (x2+ 3)(x−1− x3)0= (2x)(x−1− x3) + (x2+ 3)(−x−2− 3x2) You may stop here.= 2 − 2x4+ [−1 − 3x4− 3x−2− 9x2]= −3x−2+ 1 − 9x2− 5x42(c) h(x) = (x2+ 3)/(√x − x3)h0(x) =(x2+ 3)0(x1/2− x3) − (x2+ 3)(x1/2− x3)0(x1/2− x3)2=(2x)(x1/2− x3) − (x2+ 3)([1/2]x−1/2− 3x2)(x1/2− x3)2(3) During the summer, a group of students builds kayaks in a converted garage. Therental for the garage is $1500 for the summer, and the materials needed to build akayak cost $125. The kayaks can be sold for $275 apiece.We begin by finding a function relating the group’s profit to the number of kayaksthey sell. Let k be the number of kayaks sold. Then we haveR(k) = (price per kayak)(kayaks sold) = 275k,C(k) = (fixed cost) + (cost per kayak)(kayaks made) = 1500 + 125k,P (k) = R(k) − C(k) = 275k − (1500 + 125k) = 150k − 1500.(a) How many kayaks must the students sell to break even?0 = P (k) = 150k − 1500 ⇒ 150k = 1500 ⇒ k = 10(b) How many kayaks must the students sell to make a profit of at least $1000?1000 ≤ P(k) = 150k − 1500 ⇒ 2500 ≤ 150k ⇒ 50/3 ≤ kand so they must sell 17 kayaks to make a profit of at least $1000.(4) Specify the domain of each of these functions.(a) f(x) = x2− 2x + 6Since f(x) is a polynomial, the domain is all real numbers.(b) g(x) = (x − 3)/(x2+ x − 2)The domain of g(x) is all real numbers except for those points where the denom-inator is zero. Since we can factorx2+ x − 2 = (x + 2)(x − 1) ⇒ x = −2, x = 1we have that the domain of g(x) is all real numbers except x = −2 and x = 1.(c) h(x) =√x2− 9The square root function is defined only for nonnegative values, so we needx2− 9 ≥ 0 ⇒ x2≥ 9 ⇒ x ≥ 3 or x ≤ −3.(5) Compute the following limits. If the limit is infinite, indicate whether it is +∞ or−∞.(a) limx→3+√3x − 9Since our function is defined for x ≥ 3, we can plug in x = 3 to find that ourlimit is 03(b) limx→5+(√2x − 1 − 3)/(x − 5)limx→5+√2x − 1 − 3x − 5= limx→5+√2x − 1 − 3x − 5·√2x − 1 + 3√2x − 1 + 3= limx→5+(√2x − 1)2− 32(x − 5)(√2x − 1 + 3)= limx→5+2x − 1 − 9(x − 5)(√2x − 1 + 3)= limx→5+2(x − 5)(x − 5)(√2x − 1 + 3)=2√2 · 5 − 1 + 3=23 + 3=13(c) limx→0+x −√xSince both x and√x are defined for x ≥ 0, we can simply plug in 0 for x toobtainlimx→0+x −√x = 0 −√0 = 0.(6) A city recreation department plans to build a rectangular playground 3600 squaremeters in area. The playground is to be surrounded by a fence. Express the lengthof the fencing as a function of the length of one of the sides of the playground, drawthe graph, and estimate the dimensions of the playground requiring the least amountof fencing.Let x be the width of the playground and y be the length of the playground. Thenthe perimeter of the playground (i.e. the amount of fencing required) is given byP = 2x + 2y and the area of the playground is A = xy = 3600. Solving the equationfor area for y, we havey = 3600/x.Substituting back into our function for perimeter, we haveP (x) = 2x + 2(3600/x) = 2x + 7200/x.A graph of this function looks like:40 80 160 240 320 400 480 56080160240320400480From the graph, it appears that the minimum fencing cost will occur when x ≈ 60,in which case y = 3600/60 =
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