Midterm 1 Calculus 101 Please complete all of the following problems. Partial credit will be awarded for partial solutions, but no credit will be awarded for correct answers without accompanying work. As is stated on the syllabus, the use of notes, books and calculators are not permitted on the exam. In addition, other people are not acceptable resources. This exam is to demonstrate YOUR KNOWLEDGE, not Suzy-Q’s knowledge. You have from 7 pm on Tuesday, February 16 until 9 pm on Tuesday, February 16 to complete the following examination. My office is HB 447, and I will be there to answer questions during this time, and I will return to this room to pick up the exams at 9 pm. Except for bringing the exam to ask me a question, no exam should leave this room (HB 227) during the testing period. When you have completed the exam, please place it in the folder at the front of the classroom. When you complete the exam, please sign the following Honor Code pledge: On my honor, I have neither given nor recieved any unauthorized aid on this exam.

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Complete all of the following questions. 1. Use the formal, mathematical definition of the derivative in terms of a limit and the appropriate limit laws to show that the slope of the tangent line to a quadratic function f (x) = ax2 + bx + c at any point x is given by a linear function of x. No credit will be awarded if the limit definition of the derivative is not used. (15 pts) Solution: The slope of the tangent line to a function is given by the derivative, which measures the limiting values of the slopes of secant lines. So we compute: f (x + h) − f (x) 1 = (a(x+h)2 +b(x+h)+c−(ax2 +bx+c)) h→0 h h

f ! (x) = lim

But we can simplify, obtaining 1 f ! (x) = lim (ax2 + 2axh + ah2 + bx + bh + c − ax2 − bx − c) h→0 h 1 1 = lim (2axh + ah2 + bh) = lim (h)(2ax + ah + b) h→0 h h→0 h Because two functions which agree at all points except for x = a have the same limit at x = a, we have = lim 2ax + ah + b = lim 2ax + lim h + lim b = 2ax + 0 + b = 2ax + b h→0

h→0

h→0

h→0

where the second step follows from the sum law and the last step follows from the direct substitution law. The fuction f ! (x) = 2ax + b is a linear function, and gives the equation of the tangent line at any point on the graph of f . 2. Find the following limits if they exist. Justify your steps. (5 pts each) ) (a) limx→0 x6 cos( 1000 x Solution: Here, we don’t know whether or not the limit as x → 0 of cos( 1000 ) exists, but we do know that the cos function is bounded x from above and below by the values 1, −1 respectively. Then because x6 is always positive, it preserves inequalities. Thus −1 ≤ cos(

1000 1000 ) ≤ 1 ⇒ −x6 ≤ cos( ) ≤ x6 x x

But then limx→0 x6 = limx→0 −x6 = 0, so by the Squeeze theorem, ) = 0 as well. limx→0 x6 cos( 1000 x 2

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(b) limx→−2 xx−3x−10 2 −x−6 Solution: Begin by noting that x2 − 3x − 10 = (x − 5)(x + 2) and x2 − x − 6 = (x − 3)(x + 2). So by the fact that two functions that agree at every point except one have the same limit at that point, we know that −2 − 5 7 x2 − 3x − 10 x−5 = lim = = 2 x→−2 x − x − 6 x→−2 x − 3 −2 − 3 5 lim

where the second to last step follows by the direct substitution law. √

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(c) limx→∞ x2x3 +6−x Solution: We start by dividing the top and the bottom by the highest power in the denominator, which is x3 : ! √ 2x6 /x6 − x/x6 2x6 − x lim = lim x→∞ x3 + 6 x→∞ x3 /x3 + 6/x3 ! ! 2 − 1/x5 limx→∞ 2 − limx→∞ 1/x5 = lim = x→∞ 1 + 6/x3 limx→∞ 1 + limx→∞ 6/x3 by the quotient law, root law, and sum law √ 2 √ = = 2 1 because limx→∞ 1/xr = 0 for r rational. 3. Find the derivative. Justify your steps. (5 pts each) √ (a) g(x) = x16 + 3 x + 12 x2 + x176 Solution: g ! (x) = −6x−7 + 32 x−1/2 + x + 176x175 where here we used the power law and constant multiple law. t

e −t (b) h(t) = sin(t)−t 2 Solution: Here we need to use the quotient rule, as well as the natural log rule, and power rule.

h! (t) =

(sin(t) − t2 )(et − 1) − (et − t)(cos(t) − 2t) (sin(t) − t2 )2

There isn’t any really nice way to simplify this. It’s generally gross. 3

(c) j(u) = u2 cos(u) + u3 tan(u) Solution: here we need to use the product rule on each term in the sum, and the sum rule. j ! (u) = 2u cos(u) + u2 (− sin(u)) + 3u2 tan(u) + u3 sec2 (u) = u(2 cos(u) + u(3 tan(u) − sin(u)) + u2 sec2 u) 4. Show that there exist at least two real numbers such that five times the number is equal to the cube of the number, minus 3. Be sure to give the full statement of the theorem that you use, and explain why all of the hypotheses are fulfilled. (15 pts) Solution: Statement of the Intermediate Value Theorem: Let f be a continuous function on [a, b] and let N be any value between the values of f (a) and f (b) (assume f (a) &= f (b)). Then there exists a number c in the interval [a, b] such that f (c) = N. Suppose such a number exists; call it c. Then we know that c must satisfy the equation 5c = c3 − 3 ⇐⇒ 0 = c3 − 5c + 3. So if c exists, then c is a root of the polynomial f (x) = x3 − 5x + 3. What we want to show, then is that f has at least two real roots. As f is a polynomial, f is continuous on the reals, so in particular f is continous on any closed interval [a, b]. We need to find two intervals [a, b] and [c, d] such that the signs on f (a) and f (b) differ, and likewise for f (c) and f (d). Consider a = 0 ⇒ f (a) = 3 > 0 and b = 1 ⇒ f (1) = 1 − 5 + 3 = −1 < 0 so by the Intermediate Value Theorem (applicable because f is continuous on the closed interval [0, 1]) f has a root between x = 0 and x = 1. Now we also know that f (2) = 8 − 10 + 3 = 11 − 10 = 1 > 0 so that f (1) < 0 and f (2) > 0 implies again, by the IVT, that f has a root between x = 1 and x = 2. 5. Recall that a power function is of the form f (x) = xn . Can an even power function defined on the reals have an inverse? Why or why not? Could one restrict the domain and obtain an inverse? If so, how? (10 pts) Solution: An even power function cannot have an inverse on its whole domain. If f (x) = xn is an even power function, then n is divisible by 2, i.e. n = 2k for some real number k. But then f (x) = xn = (x2 )k and 4

then we can see that for any a, f (a) = f (−a). Thus f isn’t one-to-one, and hence cannot have an inverse on its whole domain. Alternatively, one notes that f an even power function ⇒ the graph is shaped like a parabola, and fails the horizontal line test. One could restrict the domain, by choosing {x : x > 0} or {x : x < −0}, or choosing any subdomain on which the function passes the horizontal line test. 6. Let s(t) = sin(t) + 3 cos(t) be a function describing the position of the Rice Owl at time t over the campus. (5 pts each) (a) What is the average velocity of the Owl from time t = 0 to t = π? Solution: Average velocity is given by s(π) − s(0) 0 + 3(−1) − (0 + 3) −6 = = π−0 π π (b) What is the velocity of the Owl at time t = π? Solution: Instantaneous velocity is given by v(t) = s! (t) = cos(t)− 3 sin(t). Then v(π) = cos(π) − 3 sin(π) = −1 − 3(0) = −1

(c) What is the acceleration of the Owl at the time t = 2π? Solution: The acceleration is the derivative of velocity. Thus a(t) = v ! (t) = s!! (t) = − sin(t) − 3 cos(t) = −(sin(t) + 3 cos(t)) = −s(t). Then at t = 2π, a(2π) = −(0 + 3) = −3

(d) What is the equation of the line tangent to the graph at time t = π2 ? Solution: Here we note that ( π2 , s( π2 ) = ( π2 , 1) so when we use the slope intercept form, we have y − 1 = m(x − π2 ) and we compute m = f ! ( π2 ) = cos( π2 − 3 sin( π2 ) = 0 − 3 = −3. So the equation of the line tangent to the graph at time t = π2 is y − 1 = −3(x −

π ) 2

7. (a) Find the values of c and d such that f is continuous (5 pts):  −∞ < x < −1  −x − 2 cx + d −1 ≤ x < 1 f (x) =  d 1≤x