MA131 Exam 1 Prep
https://imgur.com/7h1GgQP Pick the terms that apply to the graph above. - Constant - Increasing - Decreasing - Monotonic - Oscillatory - Unbounded - Asymptotic to the line
- Decreasing - Monotonic - Asymptotic
List the criteria needed for a limit to exist.
- Must be the same number on both sides - There must not be a discontinuity (jump) at a - Must not be infinity or negative infinity
https://imgur.com/ijdQKQL Pick the terms that apply to the graph above. - Constant - Increasing - Decreasing - Monotonic - Oscillatory - Unbounded - Asymptotic to the line
- Oscillatory - Asymptotic
RULES OF DIFFERENTIATION: d/dx (c) = ?
0
RULES OF DIFFERENTIATION: d/dx (x) = ?
1
Find the point on the graph of y = x^2 where the tangent line is parallel to the line 3x - 2y = 2.
1) Convert 3x - 2y = 2 into slope-intercept form. 3x - 2y = 2 -2y = 2 - 3x y = (3/2)x - 1 2) Since the tangent line is parallel to the given line, they both have the same slope. mTAN = 3/2 3) mTAN also equals 2x. 2x = 3/2 x = 3/4 4) To find y, plug x into y = x^2. y = (3/2)^2 = 9/16 (3/4, 9/16)
Sketch the graph of Yn+1 = 0.25Yn + 2 , Y_0 = 5
1) Find b/(1 - a) a = 0.25 b = 2 2/(1 - 0.25) = 2/0.75 = 2.67 2) Determine whether Y_0 = b/(1-a). Since 5 =/= 2.67, the graph is not constant. 3) Determine the sign of a. a = 0.25 - Positive, so the graph is monotonic. 4) Determine the size of a. |0.25| < 1 - Less than 1, so the graph is attracted to b/(1-a) = 2.67.
If y = x^4 - 5x^3 + 7, find d^2y/dx^2 when x = 3. (find f''(3)
1) Find f'(x) dy/dx = x^4 - 5x^3 + 7 = [4x^(4-1) - 5dy/dx (x^3) + dy/dx (7)] = [4x^3 - 5(3x^(3-1)) + 0] = [4x^3 - 5(3x^2)] = (4x^3 - 15x^2) 2) Find f''(x) and plug in 3. d^2y/dx^2 = (4x^3 - 15x^2) = [4(d^2y/dx^2 (x^3)) - 15(d^2y/dx^2 (x^2))] = [4(3x^2) - 15(2x)] = (12x^2 - 30x) = [12(3)^2 - 30(3)] = [12(9) - 90)] = 18
How much money can you borrow at 12% interest compounded monthly if the loan is to be paid off in monthly installments for 10 years and you can afford to pay $660 per month? [Note: (1.01)^120 = 3.3]
1) Find the difference equation. [Balance] = [Previous balance] + [Interest] - R Yn+1 = Yn + iYn - R , Y_0 = ? 2) Find r, m, i and R. Plug these in. r = 0.12 m = 12 i = r/m = 0.12/12 = 0.1 R = 660 Yn+1 = Yn + (0.1)Yn - 660 = (1 + 0.1)Yn - 660 = (1.01)Yn - 660 , Y_0 = ? 2) Solve for Y_0. Since a does not = 1, use: Yn = [b/(1 - a) + (Y_0 - (b/(1 - a)]a^n a = 1.01 b = -660 Y_0 = ? Yn = [-660/(1 - 1.01) + (Y_0 - (-660/(1 - 1.01)]1.01^n = [(-660/-0.01) + (Y_0 - (-660/-0.01)]1.01^n = 66,000 + (Y_0 - 66,000)1.01^n Since you are paying monthly for ten years, there will be 12 * 10 = 120 interest periods, or n = 120. You also want to pay the loan off after 120 periods, so Y_120 = 0. Plug n into the equation to finally solve for Y_0. 0 = 66,000 + (Y_0 - 66,000)1.01^120 = 66,000 + (Y_0 - 66,000)(3.3) = 66,000 + 3.3Y_0 - 217,800 = -151,800 + 3.3Y_0 151,800 = 3.3Y_0 Y_0 = 46,000
Suppose you want to buy a house and you know you can afford to pay $350 per month. The yearly interest for the mortgage is 8% compounded monthly. What is the maximum amount you can borrow if you want to pay off your mortgage eventually?
1) Find the difference equation. [Balance] = [Previous balance] + [Interest] - R Yn+1 = Yn + iYn - R Y_0 = ? r = 0.08 m = 12 i = r/m = 0.08/12 = 0.00667 R = 350 Yn+1 = Yn + (0.00667)Yn - 350 = (1 + 0.00667)Yn - 350 = (1.00667)Yn - 350 , Y_0 = ? 2) Graph the equation. 1 - Find b/(1 - a). a = 1.00667 b = -350 -350/(1 - 1.00667) = -350/-.00667 = 52,238.81 2 - Does b/(1 - a) = Y_0? Y_0 is unknown. 3 - Sign of a a = 1.00667 - Positive, so monotonic 4 - Size of a |1.00667| > 1 - Greater than 1, so repelled from b/(1 - a) Based on the graph, the maximum amount that can be borrowed if you want to pay off your mortgage eventually is $52,238.80.
Find the equation of the line that is tangent to the curve y = x^3 at x = 1. a) Using this tangent line, find the approximate value of 0.95^3. b) Calculate the exact value of 0.95^3.
1) Find the slope of the tangent line. y' = 3x^(3-1) = 3x^2 Plug in 1. = 3(1)^2 mTAN = 3 2) Use y - y_0 = m(x - x_0) to find the equation. y_0 = x^3 = 1^3 = 1 x_0 = 1 y - 1 = 3(x - 1) y - 1 = 3x - 3 y = 3x - 2 a) Plug in 0.95 y = 3(0.95) - 2 = 0.85 b) 0.95^3 = 0.857375 (should be close to part a if you got part a right)
In 1626, Peter Minuit purchased Manhattan Island for trinkets and cloth valued at ~$24. Suppose that this money had been invested at 7% interest compounded quarterly. How much would it have been worth by the U.S.' bicentennial year, 1976?
1) Identify the number of years between 1626 and 1976, Y_0, r, m, n, and i. Number of years: 350 Y_0 = 24 r = 0.07 m = 4 n = m(number of years) = 4(350) = 1400 i = r/m = 0.07/4 = 0.0175 2) Use the formula Yn = Y_0(1 + i)^n and plug in your values. Yn = Y_0(1 + i)^n Y_1400 = 24(1 + 0.0175)^1400 Y_1400 = 24(1.0175)^1400 Y_1400 = $850,000,000,000
Find the equation of the tangent line at the point (-2, 4) on the graph of f(x) = x^2.
1) Know that the slope of the tangent line at point (x, y) on the graph of f(x) = x^2 is determined by the formula 2x. mTAN = 2x = 2(-2) = -4 2) Plug your values into y = mx + b. y = -4x + b 4 = -4(-2) + b 4 = 8 + b b = -4 y = -4x - 4
Calculate. limx->infinity [(1/x^3 + 1)]
1) Rewrite as a complex fraction by dividing by the x with the highest power. limx->infinity [((1/x^3)/(x^3/x^3) + (1/x^3))] limx->infinity [((1/x^3)/1 + (1/x^3))] 2) Take the limit of the numerator and the denominator. limx->infinity (1/x^3) / [limx->infinity (1) + limx->infinity (1/x^3)] 0 / (1 + 0) = 0
A supermarket finds that its average daily volume of business V (in thousands of dollars) and the number of hours t that the store is open for business each day are approximately related by the formula: V = 20(1 - (100/(100 + t^2)) , 0 <= x <= 24 Find dV/dt | t = 10 and interpret what it means.
1) The problem is asking you to find the derivate of V when t = 10. dV/dt = 20d/dt(1 - 100(100 + t^2)) = 20[d/dt(1) - d/dt(100/(100 + t^2))] = 20[0 - 100(d/dt(100 + t^2)^-1)] = 20(-100)[d/dt(100 + t^2)^-1)] = -2000[(-1)(100 + t^2)^(-1-1) * d/dt (100 + t^2)] = -2000[(-1)(100 + t^2)^(-2) * (d/dt(100) + d/dt(t^2)] = 2000[(100 + t^2)^(-2) * 0 + 2t] = 2000(2t)(100 + t^2)^(-2) = 2000(2t) / (100 + t^2)^2 Plug in 10. = 2000(2(10)) / (100 + 10^2)^2 = 1 thousands per hour From hour 10 to 11, the volume of business will increase by $1,000.
Suppose that the interest rate is 5% compounded monthly. Find a formula for the amount after n months. a) Find the solution to the difference equation where Y_0 is unknown,
1) To determine the monthly rate, let: r = yearly rate (in decimal form!) m = number of times interest is given per year i = r/m Identify these variables. r = 0.05 m = 12 i = 0.05/12 = 0.004 2) The word equation for this would be: [Balance] = [Previous balance] + [Interest of previous balance] OR Yn+1 = Yn + iYn = (1 + i)Yn = (1 + 0.004)Yn = (1.004)Yn a) 1) Identify a, b and Y_0 a = 1.004 b = 0 Y_0 = ? 2) Since a does not = 1, use: Yn = [b/(1 - a) + (Y_0 - (b/(1 - a)]a^n Plug in a and b. Yn = [0/(1 - 1.004) + (Y_0 - (0/(1 - 1.004)]1.004^n = [(0/-0.004) + (Y_0 - (0/-0.004)]1.004^n = [0 + (Y_0 + 0)]1.004^n = Y_0(1.004)^n
Suppose that a savings account contains $50 and earns 4% interest, compounded annually. At the end of each year, a $3 withdrawal is made. Determine a formula (difference equation) which describes how to compute each year's balance from the previous year's balance.
1) Write a word equation to identify your variables. [Balance] = [Previous balance] + [Interest] - [Withdrawal] [Previous balance] = Y_0 = 50 [Interest] = 0.04 [Withdrawal] = -3 2) Rewrite it as: Yn+1 = Yn + aYn - W Yn+1 = Yn + (0.4)Yn - 3 Yn+1 = (1.04)Yn - 3 , Y_0 = 50
Find the derivative of f(x) = (1/x), where x does not = 0 using the limit definition.
1) Write the limit definition. limh->0 [(f(x + h) - f(x)) / h] 2) Plug in (1/x). Solve. limh->0 [(1 / (x + h)) - (1 / x) / (h / 1)] limh->0 [(1 / h)(1 / (x + h)) - (1/x)] limh->0 [(1 / h)((x / x)(1 / (x + h)) - (((x + h)/(x + h))(1 / x))] limh->0 [(1 / h)((x - (x + h) / x(x + h))] limh->0 [(1 / h)((-h) / x(x + h))] limh->0 [(1)(-1 / x(x + h))] limh->0 [(-1 / x(x + h))] 3) Plug in 0 for h limh->0 [(-1 / x(x + 0))] limh->0 (-1 / x^2)
Use limits to compute the derivative, f'(x), for the function f(x) = 9 - x^2.
1) Write the limit definition. limh->0 [(f(x + h) - f(x)) / h] 2) Plug in 9 - x^2. Solve. limh->0 [(9 - (x + h)^2) - (9 - x^2) / h] limh->0 [(9 - x^2 - 2xh - h^2) - 9 + x^2 / h] limh->0 [(-2xh - h^2) / h] limh->0 [h(-2x - h) / h] limh->0 (-2x - h) 3) Plug in 0 for h limh->0 (-2x - 0) limh->0 (-2x)
Find the derivative of f(x) = sqrt(x), where x > 0, using the limit definition.
1) Write the limit definition. limh->0 [(f(x + h) - f(x)) / h] 2) Plug in sqrt(x). Solve. limh->0 [(sqrt(x + h)) - sqrt(x) / h] Multiply the numerator and the denominator by the conjugate, (sqrt(x + h) + sqrt(x)). You get: limh->0 [h / h(sqrt(x + h) + sqrt(x))] limh->0 [1 / (sqrt(x + h) + sqrt(x))] 3) Plug in 0 for h limh->0 [1 / (sqrt(x + 0) + sqrt(x))] limh->0 [1 / (sqrt(x) + sqrt(x))] limh->0 [1 / 2sqrt(x)]
In order to buy a car, a person borrows $4,000 from the bank at 12% interest compounded monthly. The loan is to be paid off in three years with equal monthly payments. What will the monthly payments be? [Note: (1.01)^36 = 1.43]
1) [Balance] = [Previous balance] + [Interest] - R Yn+1 = Yn + iYn - R r = 0.12 m = 12 i = r/m = 0.12/12 = 0.01 Y_0 = 4000 R = ? Yn+1 = Yn + (0.01)Yn - R = (1 + 0.01)Yn - R = (1.01)Yn - R , Y_0 = 4000 2) Find the solution. Yn = [b/(1 - a) + (Y_0 - (b/(1 - a)]a^n a = 1.01 b = -R Y_0 = 4000 Yn = [-R/(1 - 1.01) + (4000 - (-R/(1 - 1.01)]1.01^n = [-R/(-0.01) + (4000 - (-R/(-0.01)]1.01^n = [(-1 * R)/(-0.01) + (4000 - (-1 * R)/(-0.01)]1.01^n = 100R + (4000 - 100R)(1.01)^n The loan is to be paid off in 3 years with monthly payments, so 3 x 12 = 36, and n = 36. Y_36 = 0 since the balance will be $0 (paid off) after 36 periods. 0 = 100R + (4000 - 100R)(1.01)^36 = 100R + (4000 - 100R)(1.43) = 100R + 5720 - 143R = -43R + 5720 -5720 = -43R R = $133.02
What does it mean when a function is differentiable? When is a function not differentiable?
A function is differentiable when a limit exists as x = a. A function is not differentiable when the graph: - is a sharp corner - is a vertical line - is a discontinuity - is an asymptote and the limits are infinite
Compare the average rate of change vs. the instantaneous rate of change.
AVERAGE RATE OF CHANGE: The average rate of change within an interval a <= x <= b (over time). Can be calculated using: f(b) - f(a) / b - a INSTANTANEOUS RATE OF CHANGE: The rate of change at a specific point, or the derivative of f(x) at x = a. f'(a) = limh->0 [f(a - h) - f(a) / h]
LIMIT THEOREMS: What does limx->a C = ?
C (constant)
What does the graph look like when Y_0 = b/(1 - a)?
Constant graph
How do derivatives apply to velocity and acceleration?
Distance: d(t) = s(t) , where d = distance t = time Velocity: v(t) = s'(t) Acceleration: a(t) = s''(t)
RULES OF DIFFERENTIATION: Constant multiple rule
If a constant is present, you can pull it to the front. d/dx[K * f(x)] = K[d/dx f(x)]
What does it mean when a graph is asymptotic?
It is attracted to the asymptote, b/(1 - a)
What does it mean when a graph is unbounded?
It is repelled from the asymptote, b/(1 - a)
True or False: Decreasing and increasing graphs are also considered monotonic.
TRUE
Describe what each of the following means when describing a graph, given a formula: - The sign of a - The size of a
The sign of a: - Positive = monotonic - Negative = oscillating The size of a: - |a| > 1 = repelled from b/(1-a) - |a| < 1 = attracted to b/(1 - a)
If h(x) = f(x) * g(x), how do you find the derivative?
Use the product rule. h'(x) = f(x) * g'(x) + f'(x) * g(x)
If h(x) = f(x) / g(x), how do you find the derivative?
Use the quotient rule. h'(x) = [g(x) * f'(x) - f(x) * g'(x)] / [g(x)]^2
What does the graph look like when a = 1?
When a = 1, we use the solution Yn = Y_0 + bn The graph forms a slanted line, where: Y_0 is the y-intercept b is the slope
Yn+1 = 1.04Yn - 3 , Y_0 = 50 Find Y_1 and Y_2.
Y_1: Y_1 = (1.04)Y_0 - 3 = (1.04)(50) - 3 = 49 Y_2: Y_2 = (1.04)Y_1 - 3 = (1.04)(49) - 3 = 47.96
What is the solution to the difference equation when a = 1?
Yn = Y_0 + bn
What is the solution to the difference equation when a does not = 1?
Yn = [b/(1 - a) + (Y_0 - (b/(1 - a)]a^n
What is the difference equation? Define each variable.
Yn+1 = aYn +b , Y_0 = c - a, b are given constants - Y_0 is the initial condition
LIMIT THEOREMS: What can you do if you have a constant K in front of f(x)? ex: limx->a K * f(x)
You can pull the constant in front. K * limx->a f(x)
LIMIT THEOREMS: What can you do if f(x) is raised to a power r? ex: limx->a [f(x)]^r
You can pull the exponent to the outside of the entire thing. [limx->a f(x)]^r If r is a positive constant
RULES OF DIFFERENTIATION: Sum/difference rule
You can take the derivates of each term separately. d/dx(x + 4) = d/dx(x) + d/dx(4)
What is the difference equation in words?
[Balance] = [Previous balance] + [Interest for period] + [Payment/Deposit]
LIMIT THEOREMS: What does limx->a x = ?
a (since y = x)
Sharon has just retired, and has $1,000,000 in her retirement account. The account will earn interest at an annual rate of 6%, compounded monthly. a) At the end of each month, Sharon will withdraw a fixed amount to cover her living expenses. Sharon wants her savings to last exactly 25 years. How much money can she withdraw each month? b) What is the maximum amount that Sharon can withdraw each month if she wants her savings to last indefinitely?
a) 1) Find the difference equation. [Balance] = [Previous balance] + [Interest] - R Yn+1 = Yn + iYn - R Y_0 = 1,000,000 r = 0.06 m = 12 i = r/m = 0.06/12 = 0.005 R = ? Yn+1 = Yn + (0.005)Yn - R = (1 + 0.005)Yn - R = (1.005)Yn - R , Y_0 = 1,000,000 2) Find the solution. Since a does not equal 1, use: Yn = [b/(1 - a) + (Y_0 - (b/(1 - a)]a^n Plug in a, b and Y_0. a = 1.005 b = -R Y_0 = 1,000,000 Yn = [-R/(1 - 1.005) + (1,000,000 - (-R/(1 - 1.005)]1.005^n = [((-1 * R)/-.005) + (1,000,000 - ((-1 * R)/-.005)]1.005^n = 200R + (1,000,000 - 200R)1.005^n Sharon will be withdrawing for 25 years x 12 months periods, which = 300 periods. Y_300 = 0 since her savings account will run out in 25 years. 0 = 200R + (1,000,000 - 200R)1.005^300 = 200R + (1,000,000 - 200R)(4.464969812) = 200R + 4,464,969.812 - 892.9939624R = -692.9939624R + 4,464,969.812 692.9939624R = 4,464,969.812 R = $6443.01 b) 1) The maximum she can withdraw if she wants her savings to last forever is basically her interest. i = r/m = 0.06/12 (0.06/12)(1,000,000) = $5,000
Compute the following limits. a) limx->3 [(x + 2)(x - 3) / (x + 5)] b) limx->3 [(x - 3) / (x^2 - 9)]
a) 1) Identify whether the denominator = 0. x + 5 3 + 5 8 does not = 0, so 2) just plug in 3. limx->3 [(x + 2)(x - 3) / (x + 5)] limx->3 [(3 + 2)(3 - 3) / (3 + 5)] limx->3 [(5)(0) / (8)] = 0 b) 1) Identify whether the denominator = 0. x^2 - 9 3^2 - 9 9 - 9 = 0 The denominator does = 0, so 2) reduce the rational function by factoring. (x - 3) / (x^2 - 9) = (x - 3) / (x - 3)(x + 3) = 1 / (x + 3) 3) Plug in 3 1 / (3 + 3) = 1/6
Suppose that a savings account contains $500 and earns 6% interest compounded annually. At the end of the year a $25 withdrawal is made. a) Determine a difference equation that describes each year's balance. b) Solve the difference equation. c) Find the balance after 100 years.
a) 1) Write a word equation to identify your variables. [Balance] = [Previous balance] + [Interest] - [Withdrawal] [Previous balance] = Y_0 = 500 [Interest] = 0.06 [Withdrawal] = -25 2) Rewrite as: Yn+1 = Yn + aYn - W Yn+1 = Yn + (0.06)Yn - W Yn+1 = (1.06)Yn - 25 , Y_0 = 500 b) 1) Identify a, b and Y_0 and whether or not a = 1. a = 1.06 b = -25 Y_0 = 500 2) Since a does not equal 1, use the solution: Yn = [b/(1 - a) + (Y_0 - (b/(1 - a)]a^n Plug in a, b and Y_0. Yn = [-25/(1 - 1.06) + (500 - (-25/(1 - 1.06)]1.06^n = [-25/(-.06) + (500 - (-25/-.06)]1.06^n = [416.66 + (500 - 416.66)]1.06^n = 416.67 + 83.33(1.06)^n c) 1) Plug in 100 for n. Yn = 416.67 + 83.33(1.06)^n = 416.67 + 83.33(1.06)^100 = 416.67 + 83.33(339.3020835) = 416.67 + 28274.04262 = $28,690.71
A person makes an initial deposit into a savings account paying 6% interest compounded annually. He plans to withdraw $120 at the end of each year. a) Find the difference equation for Yn, the amount after n years. b) How large must Y_0 be so that the money won't run out?
a) 1) Write the word equation and what it translates to. [Balance] = [Previous balance] + [Interest] - [Withdrawal] Translates to: Yn+1 = Yn + iYn - W 2) Identify r, m, i, and W. r = 0.06 m = 1 i = r/m = 0.06/1 = 0.06 W = 120 Plug these into your equation. Yn+1 = Yn + (0.06)Yn - 120 = 1.06Yn - 120 b) Analyze the graph of Yn+1 = 1.06Yn - 120 1 - Find b/(1-a) a = 1.06 b = -120 -120/(1 - 1.06) -120/-.06 = 2000 2 - Determine whether Y_0 = b/(1-a). We don't know Y_0. 3 - Determine the sign of a. a = 1.06 - Positive, so the graph is monotonic. 4 - Determine the size of a. |1.06| > 1 - Greater than 1, so the graph is repelled from b/(1 - a) Draw the horizontal line [b/(1 - a)] and the graph based on that horizontal line. Based on the graph, Y_0 must be at least $2000.
A ball is thrown up into the air. Time is measured in seconds. s(t) is the height of the ball after t seconds. s(t) = -16t^2 + 128t + 5 a) Find the average velocity of the ball during the interval [1, 2] b) Find the velocity of the ball at t = 1 c) Find the acceleration at t = 1
a) 1) f(b) - f(a) / b - a = f(2) - f(1) / 2 - 1 = 80 ft/sec The average velocity of the ball during 1-2 seconds is 80 ft/sec. b) 1) Find the derivative of s(t), then plug in 1. s'(t) = -32t + 128 = -32(1) + 128 = 96 ft/sec At 1 second, the ball is travelling at 96 ft/sec. c) a(t) = s''(t) s''(t) = -32t + 128 = -32 ft/sec^2 The acceleration of the ball at 1 second is -32 ft/sec^2 (decreases in acceleration due to gravity as it is thrown up).
LIMIT THEOREMS: How do you find the limit of a rational function? a) When the denominator does not = 0 b) When the denominator = 0
a) If the denominator does not = 0: limx->a r(x) = r(a) Just plug a into the function. b) If the denominator = 0: 1) Simplify the rational function (factor) 2) Plug x into the reduced function to find the limit
How do you find the amount after n periods for: a) Simple interest b) Compound interest
a) Yn = Y_0 + (iY_0)n b) Yn = Y_0 + (1 + i)^n Y_0 = initial amount invested/borrowed i = interest rate per period
https://imgur.com/2mwJSY6 a) At which points does the limit exist? b) At which points is the function continuous? c) At which points in the function differentiable?
a) a, b, d, f, g b) a, d, f, g c) d, f
Find f'(x) and f''(x) for each of the following: a) f(x) = 1 + 6x + x^6 b) f(x) = x^(-6) c) f(x) = (3x + 1)^4
a) f'(x) = 6 + 6x^5 f''(x) = 30x^4 b) f'(x) = -6(1/x^7) f''(x) = 42(1/x^8) c) f'(x) = 12(3x + 1)^3 f''(x) = 108(3x + 1)^2
RULES OF DIFFERENTIATION: General power rule
d/dx([g(x)^r) = r[g(x)]^(r-1) * d/dx[g(x)]
RULES OF DIFFERENTIATION: Power rule
d/dx(x^n) = nx^(n-1)
Use limits to compute the following. f'(3) when f(x) = x^2 - 2x What is the slope of the graph at x = 3?
f'(3) = limh->0 [(f(3 + h) - f(3)) / h] = limh->0 [((3 + h)^2 - 2(3 + h)) - (3^2 - 2(3))/ h] = limh->0 [((3 + h)(3 + h) - 6 - 2h - 3) / h] = limh->0 [((9 + 6h + h^2) - 6 - 2h - 3) / h] = limh->0 [(4h + h^2) / h] = limh->0 [h(4 + h) / h] = limh->0 (4 + h) = limh->0 (4 + 0) = 4 The slope of the tangent line at x = 3 is 4.
What is the limit definition and what does it describe?
limit f(x + h) - f(x) h->0 -------------- h Describes the slope of the secant line, which can be used to predict the slope of the tangent line (mTAN).
Evaluate: limx->2 sqrt(5x^3 - 15)
limx->2 (5x^3 - 15)^(1/2) [limx->2 (5x^3 - 15)]^(1/2) [limx->2 (5x^3) - limx->(15)]^(1/2) [5(limx->2 (x^3)) - 15]^(1/2) [5(limx->2 (x)^3) - 15]^(1/2) [5(limx->2 (2)^3) - 15]^(1/2) [5(limx->2 (8)) - 15]^1/2 [5(8) - 15]^(1/2) (40 - 15)^(1/2) 25^(1/2) = 5
LIMIT THEOREMS: limx->a [f(x) * g(x)] can be rewritten as...
limx->a f(x) * limx->a g(x) Same thing applies to division
LIMIT THEOREMS: limx->a [f(x) + g(x)] can be rewritten as...
limx->a f(x) + limx->a g(x) Same thing applies to subtraction
LIMIT THEOREMS: How do you find the limit of a polynomial function?
limx->a p(x) = p(a) Just plug a into the given function.
