AP CALC AB-A FINAL
False. The limit value is only equal to the function value if the function is continuous at that point. Ans: false
A limit can always be calculated by substituting the value into the function.
False. Consider f(x)=|x|. This function is not differentiable at x = 0, but is continuous there. Ans: False
A function that is not differentiable at x=c must not be continuous at x=c.
Ans: 170
Find f"(x) if f(x)= 8x^5 + 3x^3 -4x^2 +3x -7
Ans: f"(x) = secx(tan^2x+sec^2x)
Find f"(x) when f(x) = secx
Ans: f"(x) = 2cosx - x*sinx
Find f"(x) when f(x) = x*sinx
Ans: 5-2x
Find limh→0 f(x+h)−f(x)/h for f(x)=5x−x^2.
Ans: dy/dx = (sinx+1-x*cosx)/(1+sinx)^2
Find the derivative of y = x/(1+sinx)
Ans: dy/dx = (sinx+1-x*cosx)/(1+sinx)^2
Find the derivative of y=x/(1+sinx)
Ans: dy/dx = 2x*tanx - x^2*sec^2x
Find the derivative of y=x^2*tan(x).
Ans: dy/dx = 2x*tanx + x^2*sec^2x
Find the derivative of y=x^2⋅tan(x)
Ans: dy/dx = 3x^2*cotx - x^3*csc^2x
Find the derivative of y=x^3*cot(x).
Ans: dy/dx = cosx - xsinx
Find the derivative of y=xcos(x)
Ans: y - 3/5 = -4/25(x-3)
Find the equation of the line tangent to y=2x/(1+x^2) at x=3
y -2 = 3(x + 1) or y = 3x + 5 y' = 2x + 5. y' = 3, when 2x + 5 = 3 -> when x = -1. y(-1) = (-1)2+ 5(-1) + 6 = 1 - 5 + 6 = 2 tangent line; y - 2 = 3(x +1) Ans: y - 2 = 3(x-(-1)) --> y = 3x + 5.
Find the equation of the line tangent to y=x^2+5x+6 that has a slope of 3. Show your work using the equation editor or briefly describe the steps you used to get your final answer.
Let θ=arccos(4/7). Then cos⁡ cosθ=47. Since cosθ=>0, θ must be in the first quadrant as shown below. Ans: √(33)/4
Find the exact value of tan(arccos(4/7))/
(3, -5) and (1, -1) y' = 3x2-12x+9. Since the tangent line is parallel to the x-axis, the slope equals 0. y' = 0 when x = 3 or 1 or at the points (3, -5) and (1, -1). Ans: (3, -5) and (1, -1).
Find the points (x,y) on the curve y=x^3−6x^2+9x−5 where the tangent line is parallel to the x-axis. Show your work using the equation editor or briefly describe the steps you used to get your final answer.
k = -2 For the function to be continuous, limx→−1− f(x) must equal limx→−1+ f(x). limx→−1− f(x)=1−2=−1 limx→−1+ f(x)=−k−3=−1 ⇒k=−2 Ans: k = -2
Find the value for k so that the function f(x)= x^2+2x, x<−1 kx^33−3, x≥−1 is continuous. Explain how you arrived at your answer. Type your work in the text box provided using the equation editor if needed.
k = 20. For this rational function to have a limit at x = 4, the numerator must contain an (x - 4) term as a factor to cancel with the denominator. x^2+x−k=(x−4)(x+c)⇒c=5⇒k=20 If k were any other value, the function would have a vertical asymptote at x = 4. k=20
Find the value of k such that limx→4 (x^2+x−k)/(x−4) exists. Explain how you arrived at your answer. Be sure to type your steps in the text box provided using the equation editor if needed.
x = +/-.5 f(x) = 0.164 x = +/-.1 f(x) = 0.166 x = +/-.01 f(x) = 0.166665 x = +/-.001 f(x) = 0.16666665 The value of the limit is 1/6. Ans: 1/6
For each x value given below, find the value of (x-sinx)/x^3. Set your calculator to radian mode. x: ±0.5, ±0.1, ±0.01, ±0.001 f(x): ___, ___, ___, ___, ___
The line of best fit is y = 1.500x - 89.907. When x = 98, y = 57.131. Ans: 57
Given a data set, use a graphing calculator or other device to determine the line of best fit (least-squares regression line). Let x high temperature (in degrees F) and y the number of cups of lemonade sold at a local stand. Temperature(x)6871778183889195Cups Lemon.(y)1318223234444653 Use the best fit line to predict the number of cups of lemonade that will sell when the high temperature is 98.
To evaluate f(3), find the particular "piece" of the function that covers the domain including the x-value in question. Since x=3 falls into x > 1, evaluate f(3) using the "second piece". f(3) = 4 - 3^2 = -5 Ans: -5
Given f(x)= 2x+1, x≤1 4-x^2, x>1, what is f(3)?
Ans: 384
If f(x)=(2x+1)^4, then the fourth derivative of f(x) at x=0 is
If the two one-sided limits are equal, then the general limit exists and is equal to the same value. Ans: True
If limx->c- f(x) = 3 and limx->c+ f(x) = 3, then limx->c(x) = 3.
Ans: (3x^2*sec(xy)-y)/x
If sin(xy) = x^3, then dy/dx =
Ans: -3
If x^3+x*siny=1, then at the point (1,0), dy/dx =
Ans: 15sin^2(5x)cos(5x)
If y = sin^3(5x), then dy/dx=
Ans: 15sin^2(5x)cos(5x)
If y=sin^3(5x), then dy/dx=
The expression �2−6�+10=(�−3)2+1 will always have a value greater than or equal to 1 for all x. Therefore, it cannot be less than zero. Ans: There is no real solution.
Solve for x: x^2-6x+10 < 0
Ans: 12
The instantaneous rate of change of the function f(x)=x^2+4x−5 at x=4 is
Ans: y=4x+16
Which of the following is an equation for the line tangent to y=12−x^2 at x=−2?
Ans: -2x+10
Which of the following is an equation for the line tangent to y=9−x^2 at x=1?
y' = 11(2x+5)^10 * 2 = 22(2x+5)^10 y" = 220(2x+5)^9 * 2 = 440(2x+5)^9
Compute d^2y/dx^2 for y = (2x+5)^11
The derivative is (3/4)x2-3; The critical numbers are x =2 and x = -2. Using the three test intervals of (-∞ -2), (-2,2), and (2, ∞) Maximum occurs at (-2, 4)Minimum occurs at (2, -4) Ans: Maximum at (-2, 4); Minimum at (2, -4)
Determine any relative extrema using the first derivative test: f(x)=(x3/4) - 3x
1. Check continuity at x=0. A function is said to be continuous if two conditions are met. They are: the limit of the function exists and the value of the function at the point of continuity is defined and is equal to the limit of the function. The limit from the left and right equals 1 and f(0)=1. f(x) is continuous at x=0. 2. For f'(x), the limit as x approaches 0 from left equals 1 and from the right equals 0, therefore f(x) is not differentiable at x=0. We can also see there is a cusp at x=0.
Determine whether the function y= x+1, x<0 cosx, x≥0 at x=0. Explain your answer.
[8.9] - [-6.2] + [-0.7] - [3.4] = 8 - (-7) + (-1) - 3 = 11 Ans: 11
Evaluate [8.9] - [-6.2] + [-0.7] - [3.4], where [x] represents the greatest integer function
The limit is -1/2. Algebraically, divide each term by the highest power of x, which is x. This becomes limx→∞ sqrt((x^2/2)-(14/x^2)) ... Ans: -1/2
Evaluate the limit. Show all your work and explain your steps. limx→∞ sqrt(x^2−14)/3−2x
Use implicit differentiation to find the derivative and then evaluate the derivative at (6, 8). dy/dx at (6, 8) = -3/4 tangent line: y-8 = -3/4(x-6) or y=-3/4x+25/2 Ans: y - -3/4x + 25/2
Find an equation for the line tangent to the circle x^2+y^2=100 at the point (6,8) Show your work using the equation editor or briefly describe the steps you used to get your final answer.
f(x) = 1 + x^(-1) f'(x) = -x^(-2) f"(x) = 2x^(-3) = 2/x^3 Ans: f"(x) = (2)/(x^3)
Find and simplify f"(x) for f(x) = (x+1)/x Show your work using the equation editor or briefly describe the steps you used to get your final answer.
If the degrees of the numerator and denominator are the same, the horizontal asymptote equals the leading coefficient (the coefficient of the largest exponent) of the numerator divided by the leading coefficient of the denominator. In this case y = 1 Ans: y=1
Find any Horizontal Asymptotes of h(x) = (x^2-1)/(x+3)^2
Critical numbers: x=0, x=2; Distribute the x² to get f(x)= x3-3x² then differentiate to get f '(x)= 3x²-6x. Set f '(x)=0 and solve. Ans: x=0; x=2
Find any critical numbers of the function f(x)=x2(x-3).
Since y = -1 for all x except 2, dy/dx = 0. You can also show this using the quotient rule. Ans: dy/dx = 0
Find dy/dx for y = (2-x)/(x-2) (x=/ 2) Show your work using the equation editor or briefly describe the steps you used to get your final answer.
Since y = -1 for all x except 2, dy/dx = 0. You can also show this using the quotient rule. Ans: dy/dx = y' = 0.
Find dy/dx for y=(2−x)/(x−2) (x≠2) Show your work using the equation editor or briefly describe the steps you used to get your final answer.
Ans: 34
Find f"(1) if f(x) = 4x^4-3x^3+2x^2-x+1.
Ans: 24/x^4
Find f"(x) if f(x) = 4/x^2.
Ans: f"(x) = -2sinx - x*cos(x)
Find f"(x) if f(x) = x*cos(x).
Any value of x is either greater than 1 or less than 8 - and would satisfy at least one of the inequalities. If the inequality had said "AND" instead of or, the answer would be (1,8). Ans: (-∞,∞)
Solve for x: 5x-2 > 3 or x-7<1
Ans: {π/6, (5π)/6, (7π)/6, (11π)/6}
Solve the following equation on the interval [0,2π) sec(2x)−2=0.
Use quotient rule to find the derivative and then plug in the corresponding values from the table. h'(x) = [g(x)•f'(x)-f(x)•g'(x)]/[g(x)]^2 h'(2) = [(-3)(5)-(7)(2)]/(-3)^2 = -29/9 Ans: -29/9
Suppose that f(x) and g(x) are differentiable at x = 2 with the values given in the table below. f(2) f'(2) g(2) g'(2) 7 5 -3 2 h(x) = f(x)/g(x). Find h'(2)
Ans: -1
Suppose that f(x) and g(x) are differentiable for all x with the values given in the table below. f(2) f'(2) g(2) g'(2) 7. 5. -3. 2. Let h(x)= f(x)g(x). Find h'(2).
Ans: 29/49
Suppose that f(x) and g(x) are differentiable for all x with the values given in the table below. f(2) f'(2) g(2) g'(2) 7. 5. -3. 2. Let h(x)= g(x)/f(x). Find h'(2).
Ans: 20
Suppose that f(x) and g(x) are differentiable for all x with the values given in the table below. f(x). f'(x). g(x). g'(x) x=1 2. -5. 3. -1 x=2. 6. 3. 1. -4 Let h(x)= f(g(x)). Find h'(2)
Ans: sqrt(6)/4
Suppose that f(x) and g(x) are differentiable for all x with the values given in the table below. f(x). f'(x). g(x). g'(x) x=1 2. -5. 3. -1 x=2. 6. 3. 1. -4 Let h(x)=sqrt(f(x)). Find h'(2)
Ans: If f is continuous on [a,b] then f has both a minimum and a maximum on the interval
The Extreme Value Theorem states that
f(c) = 0
The Intermediate Value Theorem states that given a continuous function f defined on a closed interval [a, b] for which 0 is between f(a) and f(b), there exists a point c between a and b such that
Ans: (x^2+2x)(6x)+(3x^2+2)(3x^2-4)
The derivative of y=(x^3+2x)(3x^2−4) is
Ans: 13/(3x+5)^2
The derivative of y=2x−1/3x+5 is
y = 1/2
The equation of the horizontal asymptote for the graph of (3-e^(1/x))/(3+e^(1/x)) is
v(t) = 6t^2-42t+60 = 6(t-2)(t-5) v(t) < 0 when object is moving left so the object is moving left on the interval (2,5). Ans: the object is moving left on the interval (2,5).
The position of a particle (in feet) moving along the x-axis after t seconds is given by the following equation x(t)=2t^3−21t^2+60t−15. Determine the interval of time when the particle is moving to the left. Show your work using the equation editor or briefly describe the steps you used to get your final answer.
Ans: 1/2
The position of a particle moving along the x-axis is given by x(t)=t+cost for t≥0. The acceleration at t= 2π/3 is
Ans: π
The position of a particle moving along the x-axis is given by x(t)=t+sint for t>0. The first time the particle is at rest is at t=
Ans: 14
The position of a particle moving along the x-axis is given by x(t)=t^3+3t^2−4t+1. The average velocity of the particle from t=0 to t=3 is
Ans: 20
The position of a particle moving along the x-axis is given by x(t)=t^3+3t^2−4t+1. The velocity of the particle at t=2 is
If the limit exists, both one sided limits must be equal. If lim�→4+�(�) can be determined, then lim�→4−�(�) and lim�→4�(�) must also exist and be equal to lim�→4+�(�). Ans: True
True or False: If you know in advance that limx→4 f(x) exists, you can determine the value of the limit by knowing all the values of f(x) for all x > 4?
f(x) is a continuous function, f(1) = -3 (< 0) and f(2) = 7 (> 0). By the IVT, f(x) must take on all values between -3 and 7 on the interval [1,2]. Therefore, f(x) must equal 0 on that interval.
Use the Intermediate Value Theorem to explain why the function f(x)=x^3+3x−7 has a root on the interval [1,2]. Type your work in the text box provided using the equation editor if needed.
The limit does not exist because the two one-sided limits are not equal. limx→0+(1/x−1/|x|)=0, but limx→0−(1/x−1/|x|)=−∞ Ans: −∞
Use the concepts of one-sided limits to explain why lim�x→0 (1/x − 1/|x|) does not exist. Be sure to discuss the value of the limit on each side. Type your work in the text box provided using the equation editor if needed.
Ans: limx->1 f(x) exists.
Use the graph of f(x) above to determine which one of the following statements is true. (Graph of downward parabola w x-int at 0 and 2 and peak around 1, gap discontinuity to 2 at x=1 and y=1 for x:{2,3})
limx->1 f(x) exists.
Use the graph of f(x) above to determine which one of the following statements is true. (Graph of downward parabola w x-int at 0 and 2 and peak around 1, gap discontinuity to 2 at x=1 and y=1 for x:{2,3})
As x approaches 1, the function values approach 1. The value of the limit of a function at a point is not necessarily equal to the function value at that point. It does not matter that Ans: 1
Use the graph of f(x) to evaluate the limit. (Graph of downward parabola w x-int at 0 and 2 and peak around 1, gap discontinuity to 2 at x=1 and y=1 for x:{2,3}) limx->1 f(x) =
The limit is undefined because limx→1−f(x)≠limx→1+f(x) limx→1−f(x)=0 limx→1+f(x)=2 Results for question 5. Ans: undefined
Use the graph of f(x) to evaluate the limit. limx→1f(x)= (Graph of two parallel, downward sloping lines, gap discontinuity at x=1 from y=0 to y=1, an ar x=2 from y=1 to y=0).
As x approaches 2, the function values approach 1. The value of the limit of a function at a point is not necessarily equal to the function value at that point. It does not matter that f(2)=0. Ans: 1
Use the graph of f(x) to evaluate the limit. limx→2f(x)= (Graph of two parallel, downward sloping lines, gap discontinuity at x=1 from y=0 to y=1, an ar x=2 from y=1 to y=0).
As x approaches 1 from the left, the function approaches 0. Ans: 0
Use the graph of f(x) to evaluate the limit. limx→2f(x)= (Graph of two parallel, downward sloping lines, gap discontinuity at x=1 from y=0 to y=1, an ar x=2 from y=1 to y=0). limx->1- f(x) =
As x approaches 2 from the right, the function value is a constant 1. Ans: 1
Use the graph of f(x) to evaluate the limit. (Graph of downward parabola w x-int at 0 and 2 and peak around 1, gap discontinuity to 2 at x=1 and y=1 for x:{2,3})
d/dx (sinx/cosx) = (cosx(cosx) - (sinx)(-sinx))/(cosx)^2 = (cos^2x+sin^2x)/(cos^2x) = 1/(cos^2x) = sec^2x Ans: sec^2(x)
Use the quotient rule and the derivatives of sin x and cos x to prove that d/dxtanc=sec^2x. Show your work.
Ans: y=2x-2
What is an equation of the tangent line at x=3 if f(3)=4 and f'(3)=2?
Complete the square: �(�)=�2−6�+9−9+5=(�−3)2−4 The first term must be at least zero, so the range would be all values greater than or equal to -4. Ans: [-4,∞)
What is the range of f(x) = x^2-6x+5?
Ans: III only
Which of the following functions are differentiable for all values of x? I. f(x)=|x| II. f(x)=3√x III. f(x)=x^3
A function is one-to-one if every x value corresponds to a unique f(x) value. This is not true for I. (f(3) = f(-3)) II and III are one-to-one because no two x-values have the same function value. Ans: II and III only
Which of the following functions are one-to-one? I. y=x^2+3 II. u=x^5−2 III. y=2^(x−1)
Check each of the functions to determine if the function evaluated at x, equals the function evaluated at (-x). The only equation for which this it true is choice II. g(x)=g(-x), therefore, g(x) is even. Ans: II only
Which of the following functions is even? I. f(x) = x^2 + x II. g(x) = x^3/(x^5+x) III. h(x) = tanx
f(x) = 3/(x+1)^4
Which of the following functions is not continuous for all real numbers x?
Ans: I and II only
Which of the following functions is odd? I. f(x) = x^3 + x II. (x^2+7)/x III. sinx+cosx
Ans: graph c? straight positive slope line from l->r, y-int at ~1, axis at (1,3), then almost parabolic downward again, x-int around 2
Which of the following is a graph of f(x) = 2x+1, x≤1 4-x^2, x>1
x = 3 is a vertical line, so any perpendicular must be horizontal and have equation y = k. The horizontal line through (-2,5) is y = 5. Ans: y = 5
Write an equation for the line through (-2,5) and perpendicular to the line x = 3.
y = 3 is a horizontal line, so any perpendicular must be vertical and have equation x = k. The vertical line through (-2,5) is x = -2. Ans: x = -2
Write an equation for the line through (-2,5) and perpendicular to the line y = 3.
Ans: - cosx
d^435/dx^435 (sinx)
Ans: -cos(x)
d^435/dx^435 (sinx)
Ans: -sinx
d^89/dx^89 (cosx)
For -5<x<-4, [x]=-5, limx->-5_ [x] = -5 Ans: -5
lim x->-5+ [x]. This is the greatest integer function.
Ans: 1/6
limx->-2 ((2x+4)/(12-3x^2))
Ans: -3/5
limx->0 (3sinx)/(2x^2-5x)
Ans: 1
limx->0 (tanx/x) =
Ans: +∞
limx->0- (1/x^2)=
Ans: 5
limx->2+ 2x-3 x<2 x^2+1 x≥2
Ans: +∞
limx->2+ ((x^2+3x-4)/(x-2))
Ans: 1
limx->2- 2x-3 x<2 x^2+1 x≥2
The correct answer is 1/6. Ans: 1/6
limx->2- ((2x+4)/(12-3x^2))
Ans: 1/3
limx->3 ((2x-6)/(x^2-9))=
Ans: 0
limx->∞ ((3x^2+27)/(x^3-27))
f(x) is a continuous function connecting the points (-1,2) and (1,-2). It crosses the x-axis only once at x = m. f(x) > 0 on the interval [-1,m) and f(x) < 0 on the interval (m,1]. f(0) must be negative. Ans: -1
x: -1 0 1 f(x): 2 c -2 The function f is continuous on the closed interval [-1,1] and has values given in the table above. If �(�)=0 has only one solution, m, in the interval [-1,1] and m < 0, then a possible value of c is
