AP CALC AB GAVS

True

A function that is differentiable at x=c must be continuous at x=c.

False

A function that is not differentiable at x=c must not be continuous at x=c

(0,0) There is a cusp.

At what point (x,y) on the function f(x)=x4/5 does the derivative not exist? Explain your answer.

Step 3

Determine where the error was made when solving this limit question: lim x→25 sqrt(x)-5/25-x

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 lim x→∞ sqrt(x^2/x^2−14/x^2)3/x^2−2x/x and simplifies to lim x→∞ sqrt(1−14/x^2)/3/x−2. As x approaches ∞, any term with a power of x in the denominator approaches 0. Thus the answer is -1/2 -1/2

Evaluate the limit. Show all your work and explain your steps. lim x→∞ sqrt(x^2−14)/3−2x

Both Interior and Endpoints Points of an interval

Extrema can occur at which of the following?

Use implicit differentiation to find dy/dx = x/y. dy/dx at (3,4)= 3/4 Tangent: y-4=3/4(x-3) or y=3/4x+7/4

Find an equation for the line tangent to the hyperbola y2−x2=7 at the point (3,4). Show your work using the equation editor or briefly describe the steps you used to get your final answer.

y=2/5x+26/5

Find an equation for the line through (-3,4) and perpendicular to 5x+2y=8

y''=2sec^2x+x2sec^2xtanx

Find an expression for y'' if y=x⋅tan⁡x Show your work using the equation editor or briefly describe the steps you used to get your final answer.

y = 0

Find any Horizontal Asymptotes of g(x) = (x-3)/(x2-4)

170

Find f''(1) if f(x) = 8x^5 + 3x^3 -4x^2 +3x -7

170

Find f''(1) if f(x) = 8x^5+3x^3-4x^2+3x-7

24/x^4

Find f''(x) if f(x) = 4/x^2

f''(x) = -2sinx-x*cosx

Find f''(x) when f(x) = x * cosx

f''(x) = secx (tan^2x + sec^2x)

Find f″(x) when f(x) = secx

3

Find lim x→4 f(x) = { 7-x x<4 2, x=4 x/2 + 1 x>4 }

5-2x

Find limℎ→0 f(x+ℎ)−f(x)/ℎ for f(x)=5x^2−x^2.

y = 1

Find the Horizontal Asymptotes of f(x) = (x+2)/(x-4)

-6

Find the average rate of change of the function f(x) = x^2+4x+7 on the interval [-8, -2].

dy/dx = secx(sec^2x+tan^2x)

Find the derivative of y = tanx*secx

dy/dx = 4secx-4x*secx*tanx/(x+secx)^2

Find the derivative of y=4x/x+sec⁡x

dy/dx = 2x * tanx + x^2 * sec^2x

Find the derivative of y=x^2⋅tan⁡(x)

y = -5/3x+2/3

Find the equation of the line tangent to y=3x/x^2-4 at x=1

y=-5/3x+2/3

Find the equation of the line tangent to y=3x/x^2−4 at x=1

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) y-2 = 3(x+1) or 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.

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). (3, -5) and (1, -1)

Find the points (x,y) on the curve y=x^3−6x^2+92−5 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 = 5 For the function to be continuous, lim x→4−f(x) must equal lim x→4+ f(x). lim x→4− f(x)=16−3=13 lim x→4+ f(x)=8+k=13 ⇒k=5 k=5

Find the value for k so that the function f(x)={x^2−3, x<4 2x+k, x≥4 is continuous. Explain how you arrived at your answer. Type your work in the text box provided using the equation editor if needed.

c = -3, 4

Find the values of c so that the function f(x) = { c^2-x^2, x≤3 x+c, x>3 } is continuous for all real x

x = +/-.5 f(x) = -0.489 x = +/-.1 f(x) = -0.4995 x = +/-.01 f(x) = -0.4999995 x = +/-.001 f(x) = -0.49999996 -0.5

For each x value given below, find the value of f(x) = cos(x)-1/x^2. Set your calculator to radian mode. Then guess the value of lim x→0 (cos(x)-1/x^2). Type your work in the text box provided using the equation editor if needed.

y = 8.812x + 29.699

Given a data set, use a graphing calculator or other device to determine the line of best fit (least-squares regression line). Let x represent the hours spent studying and y the score on the test. Time studying (x): 1, 2, 4, 4, 5, 6, 6, 7, Test score (y): 41, 51, 52, 64, 73, 82, 93, 90

x^2+9x+20

Given that f(x) = x+5 and g(x) = x^2-x, find g(f(x)).

x = n * pi/2

Identify all the of vertical asymptotes of the graph of y=cot⁡(2x)

f(x) = |x-2| + 5

Identify the function graphed below:

-3x^2 cscy

If cosy = x^3, then dy/dx =

False

If f(a) exists, then lim x→a f(x) exists.

384

If f(x) = (2x+1)^4, then the fourth derivative of f(x) at x=0 is

2

If f(x) = (x+1)^2 cosx, then f'(0) =

nonesxistent

If f(x) = 4 + |x-6|, then f'(6) is

6x-1/sqrt(4x-1)

If f(x) = x sqrt(4x-1), then f'(x) =

2

If f(x)=(x+1)2cos⁡x, then f'(0) =

True

If lim x→c- f(x) = 3 and lim x→c+ f(x) = 3, then lim x→c f(x) = 3 exists.

-3

If x^3+x * siny = 1, then at the point (1,0), dy/dx =

15sin^2(5x)cos(5x)

If y = sin^3(5x), then dy/dx =

Five

Let f be the function f(x)=sin(x2+1). How many relative extrema does f have on the interval -2 ‹ x ‹ 2?

{x: -10 ≤ x ≤ 6}

Let f(x) = sqrt(4-x) and g(x) = sqrt(6-x). What is the domain of f(g(x))?

(5/2 < x < 11/2)

Solve for x: |2x-8| < 3

[-12,2]

Solve for x: |x+5| ≤ 7

{pi/4, 3pi/4, 5pi/4, 7pi/4}

Solve the following equation on the interval [0, 2pi). 2sin^2x-1=0

(h'(x)=f'(g(x))⋅g'(x)\)ℎ′(5)=f′(g(5))⋅g′(5)=f′(3)⋅g′(5) We do not have enough information. h'(5) cannot be computed without f'(3). Not enough information

Suppose that f(5) = f'(5) = g(5) = g'(5) = 3. Is this enough information to compute h'(5) where h(x) = f(g(x))? Why or why not?

29/49

Suppose that f(x) and g(x) are differentiable at x = 2 with the values given in the table below. f(2): 7 f'(2): 5 g(2): -3 g'(2): 2 Let h(x) = g(x)/f(x). Find h'(2)

-1

Suppose that f(x) and g(x) are differentiable at x = 2 with the values given in the table below. f(2): 7 f′(2): 5 g(2): -3 g′(2): 2 ℎ(x)=f(x)g(x). Find h'(2)

-27

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 2-53-1x=2 631-4 Let ℎ(x)=(g(x))^3. Find ℎ′(1)

10

The average rate of change of the function f(x) = x^2+4x-5 over the closed interval [1,5] is

(x^3+2x)(6x)(3x^2+2)(3x^2-4)

The derivative of y=(x^3+2x)(3x^2−4) is

-25/(2x-7)^2

The derivative of y=3x+2/2x−7 is

12

The instantaneous rate of change of the function f(x) = x^2 +4x - 5 at x = 4 is

0

The maximum value of f(x) = x3 + 3x² - 9x - 2 on the interval [0, 2] is

To find the velocity of the particle at the instant the accelerationis 0, we will first need to find the acceleration function from the position function. This can be done by taking the derivative twice from the original, position function. Then we will need to set the acceleration function equal to 0. Once we have the value at which the acceleration function is equal to 0, then we will need to plug that value back into the first derivative, which will be the velocity function. The work goes as follows: f(x)=x^3−9x^2+5x−7 f′(x)=3x^2−18x+5 - velocity function f''(x) = 6t-18 We will then set 6t-18 equal to 0 in which we get 3. We will plug 3 back in to the velocity function: 3(3)2−18(3)+5 in which we get -22 feet. -22 ft/sec

The position of a particle (in feet) moving along the x-axis after t seconds is given by the following equation x(t)=t^3−9t^2+5t−7. Find the velocity of the particle at the instant the acceleration is 0. Show your work using the equation editor or briefly describe the steps you used to get your final answer.

-2/pi

The position of a particle moving along the x-axis is given by x(t)=cos(t). The average velocity of the particle from t=0 to t=π is

4

The position of a particle moving along the x-axis is given by x(t)=t^3−5t^2−8t+2 for t≥0. The particle is at rest is at t =

4

The position of a particle moving along the x-axis is given by x(t)=t^3−5t^2−8t+2 for t≥0. The particle is at rest is at t=

-4

The position of a particle moving along the x-axis is given by x(t)=t^3−5t^2−8t+2. The acceleration at t=1 is

False

True or False: if lim x→a f(x) exists, then f(a) exists.

Local maximum at x = 1/3 Local minimum at x = 5

Use a calcuator to find the x-coordinates of the local extrema of f(x)=x^3−8x^2+5x+1

lim theta→pi (thetasectheta) lim theta→pi theta/cos(theta) = pi/-1 = -pi -pi

Use an analytic technique to determine lim theta→pi(thetasectheta Be sure to type your steps in the text box provided using the equation editor if needed.

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. lim x→3+ [x] = 3 but lim x→3- [x] = 2 DNE

Use the concepts of one-sided limits to explain why lim x→3 [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. (Note: [x]=[x]=int(x)) This is the greatest integer function.

lim x→1 f(x) exists.

Use the graph above to determine which one of the following statements is true. (1 looks like a parabola with y=1 line)

1

Use the graph of f(x) to evaluate the limit. lim x→1 f(x)= (1 looks like a parabola with y=1 line)

undefined

Use the graph of f(x) to evaluate the limit. lim x→1 f(x)= (2 lines with 2 points in between)

2

Use the graph of f(x) to evaluate the limit. lim x→1+ f(x) =

1

Use the graph of f(x) to evaluate the limit. lim x→2 f(x) =

0

Use the graph of f(x) to evaluate the limit. lim x→2- f(x) =

y = sinx * sinx use product rule. y'=2sinxcosx=sin(2x)

Use the product rule and a trigonometric identity to prove d/dx(sin^2x)=sin⁡(2x). Show your work.

Too many steps: But -> 1/cos^2x = sec^2x High Order Derivatives Quiz #10

Use the quotient rule and the derivatives of sin x and cos x to prove that d/dx tan⁡x=sec^x. Show your work.

[-4, infinity)

What is the range of f(x) = x^2-6x+5?

[0,4]

What is the range of f(x)= sqrt((16-(x-1)^2)

III only

Which of the following functions are differentiable for all values of x? I. f(x)=|x| II. f(x)=cube root(x) III. f(x)=x^3

f(x) = 3/(x+1)^4

Which of the following functions is not continuous for all real numbers x?

y=-2x+10

Which of the following is an equation for the line tangent to y = 9-x^2 at x=1

y = -x + pi/2 +1

Which of the following is an equation of the tangent line to y=sin⁡x+cos⁡x at x = pi/2

y = 0

Which of the following is an equation of the tangent line to y=sin⁡x+tan⁡x at x=pi?

x=1

Which of the following points of discontinuity of f(x) = (x(x-1)^2(x-2)^2(x-3)^2)/(x(x-1)^3(x-2)^2(x-3))

-sinx

d^89/dx^89 (cosx) =

1/6

lim x→-2 (2x+4/12-3x^2) =

7

lim x→0 ((sin(7x)/x) =

1

lim x→0 (tanx/x)

0

lim x→0+ = This is the greatest integer function. (Note: [x] = [x] = int(x))

- infinity

lim x→2- (x^2+3x-4/x-2)

1/3

lim x→3 (2x-6/x^2-9)

-1

lim x→3- (|x-3|/x-3)

0

lim x→infinity (3^-x)/(3^x)

-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 f(x)=0 has only one solution, m, in the interval [-1,1] and m < 0, then a possible value of c is