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Worksheet # 1: Functions and inverse functions
- Give the domain and ranges of the following functions.
(a) f (x) = (^) x 2 x++1x− 2 Solution: a) The domain is {x : x 6 = −2 and x 6 = 1} and the range is all real numbers, R = (−∞, ∞). (b) g(t) = √t (^21) − 1 Solution: b) The domain {t : − 1 < t < 1 } = (− 1 , 1) and the range is {t : t ≥ 1 } = [1, ∞).
- If f (x) = 5x + 7 and g(x) = x^2 , find f ◦ g and g ◦ f. Are the functions f ◦ g and g ◦ f the same function? Solution: The two functions, f ◦ g and g ◦ f , are not the same.
- Let f (x) = 2 + (^) x+3^1. Determine the inverse function of f , f −^1. Give the domain and range of f and
the inverse function f −^1. Verify that f ◦ f −^1 (x) = x.
- Consider the function whose graph appears below.
1
1
x
y=f(x) (a) Find f (3), f − (^1) (2) and f − (^1) (f (2)).
(b) Give the domain and range of f and of f −^1. (c) Sketch the graph of f −^1.
- Let f (x) = x^2 + 2x + 5. Find the largest value of a so that f is one to one on the interval (−∞, a]. Let g be the function f with the domain (−∞, a]. Find the inverse function g−^1. Give the domain and range of g−^1. Solution: See Written Assignment 1 Solutions
- True or False:
(a) Every function has an inverse. Solution: False. A function must be one-to-one to have an inverse (b) If f ◦ g(x) = x for all x in the domain of g, then f is the inverse of g. Solution: False. The functions must also satisfy g ◦ f (x) = x for all x in the domain of f to be inverses (c) If f ◦ g(x) = x for all x in the domain of g and g ◦ f (x) = x for all x in the domain of f , then f is the inverse of g. Solution: True. (d) If f (x) = 1/(x + 2)^3 and g is the inverse function of f , then g(x) = (x + 2)^3. Solution: False. If the functions where to be inverses we would have g ◦ f (x) = x and f ◦ g(x) = x, but these are not true. (e) The function f (x) = sin(x) is one to one. Solution: False. The sine function doesn’t satisfy the horizontal line test.
(f) The function f (x) = 1/(x + 2)^3 is one to one. Solution: True.
- Find the slope, x-intercept, and y-intercept of the line 3x − 2 y = 4. Solution: The slope is 32 , the x-intercept is ( 43 , 0) and the y-intercept is (0, −2)
- Let f be a linear function with slope m with m 6 = 0. What is the slope of the inverse function f −^1. Solution: Since f is a linear function, we have that f (x) = mx + b for some constants m and b. To find the inverse of f : y = mx + b x = my + b my = x − b
y =
m
x −
b m Therefore f −^1 (x) = (^) m^1 x − (^) mb and the slope is (^) m^1.
- A ball is thrown in the air from ground level. The height of the ball in meters at time t seconds is given by the function h(t) = − 4. 9 t^2 + 30t. At what time does the ball hit the ground (be sure to use the proper units)? Solution: When the ball hits the ground the height of the ball is zero, so
h(t) = − 4. 9 t^2 + 30t = 0
t(− 4. 9 t + 30) = 0 This means that either t = 0 or − 4. 9 t + 30 = 0. When t = 0 the ball is just being thrown. This is not the time wanted. The other option gives t = (^430). 9. Hence the ball hits ground after approximately 6 .1224 seconds.
- We form a box by removing squares of side length x centimeters from the four corners of a rectangle of width 100 cm and length 150 cm and then folding up the flaps between the squares that were removed. a) Write a function which gives the volume of the box as a function of x. b) Give the domain for this function. Solution: See Written Assignment 1 Solutions
(a)
3 cos(x) + 2 tan(x) cos^2 (x) = 0 Solution: √ 3 cos(x) + 2 (^) cos(sin(xx)) cos^2 (x) = cos(x)(
3 + 2 sin(x)) = 0. Setting cos(x) = 0 and sin(x) = −
√ 3 2 , we find the solutions are π 2 , 32 π , 43 π , 53 π. (b) 3 cot^2 (x) = 1 Solution: 3
cos^2 (x) sin^2 (x)
3 cos^2 (x) = 1 − cos^2 (x) 4 cos^2 (x) = 1
cos(x) = ±
gives us solutions of π 3 , 23 π , 43 π , 53 π. (c) 2 cos(x) + sin(2x) = 0 Solution: 2 cos(x) + sin(2x) = 2 cos(x) + 2 sin(x) cos(x) = 2 cos(x)(1 + sin(x)) = 0. Setting cos(x) = 0 and sin(x) = −1, we find the solutions are π 2 , 32 π.
- A function is said to be periodic with period T if f (x) = f (x + T ) for any x. Find the smallest, positive period of the following trigonometric functions. Assume that ω is positive.
(a) | sin t| Solution: T = π (b) sin(3t). Solution: T = 2π/ 3 (c) sin (ωt) + cos (ωt). Solution: The period of sin(t) + cos(t) is 2π. Thus the period of sin(ωt) + cos(ωt) is 2π/ω. (d) tan^2 (ωt). Solution: The period of tan^2 (t) is π. Then the period of tan^2 (ωt) is π/ω.
- Find a quadratic function p(x) so that the graph p has x-intercepts at x = 2 and x = 5 and the y-intercept is y = −2. Solution: Since p has x-intercepts, x = 2 and x = 5, we can write p(x) = a(x − 2)(x − 5) Since the y-intercept is −2, −2 = p(0) = 10a. Hence a = − 15 and so p(x) = − 15 (x − 2)(x − 5) is the solution.
- Find the exact values of the following expressions. Do not use a calculator.
(a) tan−^1 (1) Solution: Solving tan−^1 (1) is equivalent to solving tan(θ) = 1 for θ in [− π 2 , π 2 ]. Our solution is π
(b) tan(tan−^1 (10)) (c) sin−^1 (sin(7π/3)) (d) tan(sin−^1 (0.8))
- Give a simple expression for sin(cos−^1 (x)).
Solution: Using x = cos(θ), we have that sin(cos−^1 (x)) = sin(θ). Since cos(θ) = (^) hypotenuseadjacent = x, we can sketch a picture and solve for θ using the Pythagorean Theorem. If θ is the bottom left angle, we see that sin(θ) =
1 − x^2 , where we chose the positive root because the range of cos−^1 (x) is [0, π] and on this domain sine is positive.
x
(^11) − x^2
Hence sin(cos−^1 (x) =
1 − x^2
- Let f be the function with domain [π/ 2 , 3 π/2] with f (x) = sin(x) for x in [π/ 2 , 3 π/2]. Since f is one to one, we may let g be the inverse function of f. Give the domain and range of g. Find g(1/2). Solution: The domain of g is [− 1 , 1] and the range of g is [π/ 2 , 3 π/2]. Now if g(1/2) = θ, then sin(θ) = 12. This occurs at θ = π 6 and 56 π. Since we know the domain of g is [π/ 2 , 3 π/2], the answer is g( 12 ) = 56 π.
- Consider the function f (x) = 1 + ln(x). Determine the inverse function of f. Give the domain and range of f and of the inverse function f −^1.
- Suppose that a population doubles every two hours. If we have one hundred critters at 12 noon, how many will there be after 1 hour? after 2 hours? How many were there at 11am? Give a formula for the number of critters at t hours after 12 noon. Solution: We use the fact that ex^ and ln(x) are inverses to get 4x^ = eln(
x) , and we use properties of logarithms to get eln( x) = eln(4)x.
- Let f be the function f (x) = 4x. Write the function f in the form f (x) = ekx.
- Let f be the function f (x) = 5 · 3 x. Write the function in the form Aekx.
- Suppose that f is a function of the form f (x) = Aekx. If f (2) = 20 and f (5) = 10, will we have k > 0 or k < 0? Find A and k so that f (2) = 20 and f (5) = 10. Solution: We know that f (x) is of the form f (x) = Aekx. Also, we have that k > 0 if and only if f (x) is increasing, and k < 0 if and only if it is decreasing. Since f (2) = 20 and f (5) = 10, f (x) is getting smaller as x increases, so f (x) is decreasing, and therefore k < 0. To solve for A and k exactly, note that we’re given f (2) = Ae^2 k^ = 20 and f (5) = Ae^5 k^ = 10. If we multiply each side of the second equation by 2, we get 2Ae^5 k^ = 20, so we can set it equal to Ae^2 k^ and solve for k:
2 Ae^5 k^ = Ae^2 k 2 e^3 k^ = 1 (division by Ae^2 k)
e^3 k^ =
ln(e^3 k) = ln(1/2) 3 k = − ln(2) (because ln(1/2) = ln(2−^1 ) = − ln(2))
k =
− ln(2) 3
And we can plug this k value into Ae^2 k^ = 20 to get our answer for A:
Ae^2 ·−.^23 = 20
A =
e^2 ·−.^23
- Let f be the function given by f (x) = tan(x) with x in the interval (π/ 2 , 3 π/2). Find f −^1 (0) and f −^1 (
Sketch the graphs of f and f −^1. Give a formula which expresses f −^1 in terms of the function tan−^1 (or arctan). Solution: We have f −^1 (x) = π + arctan(x), so f −^1 (0) = π + arctan(0) = π and
f −^1 (
- = π + arctan(
- = π + π/3 = 4π/ 3.
Worksheet # 4: Limits: A Numerical and Graphical Approach, the
Limit Laws
- Comprehension check:
(a) In words, describe what “ lim x→a f (x) = L” means.
(b) In words, what does “ lim x→a f (x) = ∞” mean?
(c) Suppose lim x→ 1
f (x) = 2. Does f (1) = 2?
(d) Suppose f (1) = 2. Does lim x→ 1 f (x) = 2?
- Let p(t) denote the distance (in meters) to the right of the origin of a particle at time t minutes after noon. If p(t) = p(t) = t^3 − 45 t, give the average velocity on the intervals [2, 2 .1] and [2, 2 .01]. Use this information to guess a value for the instantaneous velocity of particle at 12:02pm.
- Consider the circle x^2 + y^2 = 25 and verify that the point (− 3 , 4) lies on the circle. Find the slope of the secant line that passes through the points with x-coordinates −3 and − 3 .1. Find the slope of the secant line that passes through the points with x-coordinates −3 and − 2 .99. Use this information to guess the slope of the tangent line to the circle at (− 3 , 4). Write the equation of the tangent line and use a graphing calculator to verify that you have found the tangent line.
- Compute the value of the following functions near the given x−value. Use this information to guess the value of the limit of the function (if it exists) as x approaches the given value.
(a) f (x) = 4 x
(^2) − 9 2 x− 3 ,^ x^ =^
3 2 (b) f (x) = (^) |xx| , x = 0
(c) f (x) = sin(2 x x), x = 0 (d) f (x) = sin(π/x), x = 0
Solution:
(a) (b) f (x) = (^) |xx| , x = 0
x = 0. 1 , f (x) = 1 x = 0. 01 , f (x) = 1 x = 0. 001 , f (x) = 1
x = − 0. 1 , f (x) = − 1 x = − 0. 01 , f (x) = − 1 x = − 0. 001 , f (x) = − 1
This indicates that limx→ 0 − = −1, but limx→ 0 + = 1. The left and right limits are not equal, and so the limit does not exist.
(c) f (x) = sin(2 x x), x = 0
x = 0. 1 , f (x) = 1. 98 x = 0. 01 , f (x) = 1. 999 x = 0. 001 , f (x) = 1. 99999
x = − 0. 1 , f (x) = 1. 98 x = − 0. 01 , f (x) = 1. 999 x = − 0. 001 , f (x) = 1. 99999
As x approaches 0 from both the positive and negative directions, f (x) approaches 2, so lim x→ 0
sin(2x) x
(d)
- Let f (x) =
2 x + 2 if x > − 2 a if x = − 2 kx if x < − 2
. Find k and a so that lim x→− 2 f (x) = f (−2).
- In the theory of relativity, the mass of a particle with velocity v is:
m =
m 0 √ 1 −
v^2 c^2
where m 0 is the mass of the particle at rest and c is the speed of light. What happens as v → c−?
Worksheet # 5: Continuity
- Comprehension check:
(a) Define what it means for f (x) to be continuous at the point x = a. What does it mean if f (x) is continuous on the interval [a, b]? What does it mean to say f (x) is continuous? (b) There are three distinct ways in which a function will fail to be continuous at a point x = a. Describe the three types of discontinuity. Provide a sketch and an example of each type. (c) True or false? Every function is continuous on its domain. (d) True or false? The sum, difference, and product of continuous functions are all continuous. (e) If f (x) is continuous at x = a, what can you say about lim x→a+
f (x)?
- Using the definition of continuity and properties of limits, show that the following functions are con- tinuous at the given point a. (a) f (x) = π, a = 1
(b) f (x) =
x^2 + 3x + 1 x + 3
, a = − 1
(c) f (x) =
x^2 − 9, a = 4
- Give the intervals of continuity for the following functions.
(a) f (x) = x + 1 x^2 + 4x + 3 (b) f (x) =
x x^2 + 1 (c) f (x) =
2 x − 3 + x^2
(d) f (x) =
x^2 + 1 if x ≤ 0 x + 1 if 0 < x < 2 −(x − 2)^2 if x ≥ 2
- If limh→ 3 (hf (h)) = 15, can you find limh→ 3 f (h)? Use the limit laws to justify your answer.
- Let c be a number and consider the function f (x) =
cx^2 − 5 if x < 1 10 if x = 1 1 x
− 2 c if x > 1
(a) Find all numbers c such that lim x→ 1 f (x) exists. (b) Is there a number c such that f (x) is continuous at x = 1? Justify your answer.
- Find parameters a and b so that f (x) =
2 x^2 + 3x if x ≤ − 4 ax + b if − 4 < x < 3 −x^3 + 4x^2 − 5 if 3 ≤ x
is continuous.
- Suppose that f (x) and g(x) are continuous functions where f (2) = 5 and g(6) = 1. Compute the following:
(a) lim x→ 2
[f (x)]^2 + x 3 x + 2
(b) lim x→ 6
g(x) + 4x f
( (^) x 3
− g(x)
- Suppose that: f (x) =
x− 6 |x− 6 | for^ x^6 = 6, 1 for x = 6 Determine the points at which the function f (x) is discontinuous and state the type of discontinuity.
Worksheet # 7: The Intermediate Value Theorem
- (a) State the Intermediate Value Theorem. (b) Show that f (x) = x^3 + x − 1 has a zero in the interval [0, 1].
- Let f (x) = ex/(ex^ − 2). Show that f (0) < 1 < f (ln(4)). Can you use the intermediate value theorem on the interval [0, ln(4)] to conclude that there is a solution of f (x) = 1? Solution: No, the function f is not continuous on the interval [0, ln(4)]. Can you find a solution to f (x) = 1? Solution: No. If we try to solve ex ex^ − 2
we obtain the equation ex^ = ex^ − 2 or 0 = −2 which is not true.
- Use the Intermediate Value Theorem to find an interval of length 1 in which a solution to the equation 2 x^3 + x = 5 must exist.
- Show that there is some a with 0 < a < 2 such that a^2 + cos(πa) = 4.
- Show that the equation xex^ = 2 has a solution in the interval (0, 1). Solution: The function xex^ is a product of the continuous functions x and ex^ and hence continuous. We have f (0) = 0 and f (1) = e ≈ 2 .78. Since 0 < 2 < e, the intermediate value theorem tells us there is a solution to the equation f (x) = 2. Determine if the solution lies in the interval (0, 1 /2) or (1/ 2 , 1). Continue to find an interval of length 1/8 which contains a solution of the equation xex^ = 2.
- Find an interval for which the equation ln(x) = e−x^ has a solution. Use the intermediate value theorem to check your answer.
- Let f (x) =
0 if x ≤ 0 1 if x > 0
be a piecewise function.
Although f (−1) = 0 and f (1) = 1, f (x) 6 = 1/2 for all x in its domain. Why doesn’t this contradict to the Intermediate Value Theorem?
- Prove that x^4 = −1 has no solution.
- Determine if the following are true or false.
(a) If f is continuous on [− 1 , 1], f (−1) = −2 and f (1) = 2, then f (0) = 0. (b) If f is continuous on [− 1 , 1], f (−1) = −2 and the equation f (x) = 1 has no solution, then there is no solution to f (x) = 0. (c) If f is continuous on [− 1 , 1], f (−1) = 1, and the equation f (x) = 1 has no solution, then there is no solution to f (x) = 2. (d) If f is a function with domain [− 1 , 1], f (−1) = −2, f (1) = 2, and for each real number y, we can find a solution to the equation f (x) = y, then f is continuous.
- (Review) A particle moves along a line and its position after time t seconds is p(t) = 3t^3 + 2t meters to the right of the origin. Find the instantaneous velocity of the particle at t = 2.
Worksheet # 8: Review for Exam I
- Find all real numbers of the constant a and b for which the function f (x) = ax + b satisfies:
(a) f ◦ f (x) = f (x) for all x. (b) f ◦ f (x) = x for all x.
- Simplify the following expressions.
(a) log 5 125 (b) (log 4 16)(log 4 2) (c) log 15 75 + log 15 3 (d) logx(x(logy yx)) (e) logπ (1 − cos x) + logπ (1 + cos x) − 2 logπ sin x
- Suppose that tan(x) = 3/4 and −π < x < 0. Find cos(x), sin(x), and sin(2x).
- (a) Solve the equation 3^2 x+5^ = 4 for x. Show each step in the computation.
(b) Express the quantity log 2 (x^3 − 2) +
log 2 (x) − log 2 (5x) as a single logarithm. For which x is the resulting identity valid?
- Calculate the following limits using the limit laws. Carefully show your work and use only one limit law per step.
(a) lim x→ 0 (2x − 1) Solution: lim x→ 0 (2x − 1) = lim x→ 0 (2x) − lim x→ 0
lim x→ 0 (2x − 1) = 2 lim x→ 0 (x) − 1 lim x→ 0 (2x − 1) = 2(0) − 1 = − 1
(b) lim x→ 0
x + 4 − 2 x Solution: lim x→ 0
x + 4 − 2 x
= lim x→ 0
x + 4 − 2 x
x + 4 + 2 √ x + 4 + 2
lim x→ 0
x + 4 − 2 x
= lim x→ 0
x + 4 − 4 x
x + 4 + 2
lim x→ 0
x + 4 − 2 x
= lim x→ 0
x x
x + 4 + 2
lim x→ 0
x + 4 − 2 x
= lim x→ 0
x + 4 + 2
lim x→ 0
x + 4 − 2 x
lim x→ 0
x + 4 + 2)
lim x→ 0
x + 4 − 2 x
lim x→ 0
(x) + 4 + 2)
lim x→ 0
x + 4 − 2 x
- Determine if the following limits exist and calculate the limit when possible.
lim x → a
(x − a)(x + a) x − a
lim x → a(x + a) = 8 2 a = 8
Therefore a = 4 makes f continuous.
- Complete the following statements:
(a) A function f (x) passes the horizontal line test, if the function f is ................ Solution: one-to-one (b) If lim x→a f (x) and lim x→a g(x) exist, then
lim x→a
f (x) g(x)
lim x→a f (x) lim x→a g(x)
provided ..................
Solution: lim x→a g(x) 6 = 0
(c) lim x→a+
f (x) = lim x→a−
f (x) = L if and only if ................... Solution: lim x→a f (x)exists
(d) Let g(x) =
x if x 6 = 2 1 if x = 2
be a piecewise function.
The function g(x) is NOT continuous at x = 2 since ................................. Solution: lim x→ 2 g(x) 6 = g(2).
(e) Let f (x) =
x^2 if x < 0 1 if x = 0 x if x > 0
be a piecewise function.
The function f (x) is NOT continuous at x = 0 since ................................. Solution: 0 = lim x→ 0 −
f (x) = lim x→ 0 +
f (x) 6 = f (0) = 1
- If g(x) = x^2 + 5x^ − 3, use the Intermediate Value Theorem to show that there is a number a such that g(a) = 10. Solution: Since 3 = g(1) < 10 < g(2) = 26 and g is a continuous function, the Intermediate Value Theorem says that there is some number c ∈ (1, 2) such that g(c) = 10
Worksheet # 9: The Derivative
- Comprehension check:
(a) What is the definition of the derivative f ′(a) at a point a? Solution: Students should learn the definition as in the textbook. The derivative f ′(a) gives the slope of the line that is tangent to the graph of f at the point a. (b) What is the geometric meaning of the derivative f ′(a) at a point a? (c) True or false: If f (1) = g(1), then f ′(1) = g′(1)? Solution: False. Consider f (x) = x^2 and g(x) = x.
- Consider the graph below of the function f (x) on the interval [0, 5].
(a) For which x values would the derivative f ′(x) not be defined? (b) Sketch the graph of the derivative function f ′.
- Find a function f and a number a so that the following limit represents a derivative f ′(a).
lim h→ 0
(4 + h)^3 − 64 h Solution: According to the definition of the derivative, an obvious choice is f (x) = x^3 and a = 4.
- Let f (x) = |x|. Find f ′(1), f ′(0) and f ′(−1) or explain why the derivative does not exist.
Solution: Looking at the graph of |x|, we can see the slope of the tangent lines to get f ′(1) = 1 and f ′(−1) = −1, but f ′(0) does not exist as the slope is different on each side of x = 0. This means left and right limits are not equal.
- Find the specified derivative for each of the following.
(a) If f (x) = 1/x, find f ′(2). (b) If g(x) =
x, find g′(2). (c) If h(x) = x^2 , find h′(s).
- Find A and B so that the limit
lim x→ 1
x^2 + 2x − (Ax + B) (x − 2)^2 is finite. Give the value of the limit.
Worksheet # 10: The Derivative as Function, Product and Quotient
Rules
- Comprehension check:
(a) True or false: If f ′(x) = g′(x) for all x, then f = g? (b) True or false: If f (x) = g(x) for all x, then f ′^ = g′? (c) True or false: (f + g)′^ = f ′^ + g′ (d) True or false: (f g)′^ = f ′g′ (e) How is the number e defined? (f) Are differentiable functions also continuous? Are continuous functions also differentiable?
- Show by way of example that, in general,
d dx
(f · g) 6 =
df dx
dg dx and
d dx
f g
df dx dg dx
- Calculate the derivatives of the following functions in the two ways that are described.
(a) f (r) = r^3 / 3 i. using the constant multiple rule and the power rule ii. using the quotient rule and the power rule Which method should we prefer? (b) f (x) = x^5 i. using the power rule ii. using the product rule by considering the function as f (x) = x^2 · x^3 (c) g(x) = (x^2 + 1)(x^4 − 1) i. first multiply out the factors and then use the power rule ii. by using the product rule
- State the quotient and product rule and be sure to include all necessary hypotheses.
- Compute the first derivative of each of the following:
(a) f (x) = (3x^2 + x)ex
(b) f (x) =
x x − 1
(c) f (x) = ex 2 x^3
(d) f (x) = (x^3 + 2x + ex)
x − 1 √ x
(e) f (x) = 2 x 4 + x^2
(f) f (x) =
ax + b cx + d
(g) f (x) =
(x^2 + 1)(x^3 + 2) x^5 (h) f (x) = (x − 3)(2x + 1)(x + 5)
- Let f (x) = (3x − 1)ex. For which x is the slope of the tangent line to f positive? Negative? Zero?
