Name___________ Rate-in Rate-out Super FRQ (Calculator Active) 1) Coal is burning in a furnace, thus exhausting the resource. The rate at which coal is burning, measured in pounds per hour, is given by . At t = 2 hours a worker starts supplying additional coal (t) 8sin ( )B = t

2π into the furnace. The rate at which coal is being added, measured in pounds per hour, is given by

. The worker stops adding coal at t = 5 hours. At t = 0 there are 500 pounds of (t) 11 10cos( )S = + 254πt

coal in the furnace. (a) Find the total amount of coal, in pounds, added by the worker. (b) Is the amount of coal in the furnace increasing or decreasing at t = 4 hours? Explain. (c) Find S’(3). Explain, with units, the meaning of this in the context of the problem. (d) Write a piecewise function A(t), that gives the total amount of coal in the furnace at any time

. 0 ≤ t ≤ 5 (e) Find the minimum and maximum amount of coal, in pounds, from time t = 0 to t = 5 hours and the times that they occur. (f) Find the average rate of coal burned from t = 0 to t = 8 hours. (g) After t = 5 hours, the coal continues to burn until there are 0 pounds of coal remaining. Write an equation involving one or more integral expressions, that gives the time t = k, that the coal is exhausted.

Name____________ Position Velocity Acceleration Super FRQ (Calculator Active) 2) A particle moves along the x-axis so that its velocity at time t is given by :

At time t = 0, the initial position of the particle is x = 5. a. Find the acceleration of the particle at time t = 5 b. Is the particle speeding up or slowing down at time t = 5.1? Justify your answer. c. Find all times on the interval 0 < t < 5, that the particle is at rest. d. Write an expression involving an integral that gives the position x(t). Find the position of the particle at t = 2. e. Calculate the total distance traveled by the particle from t = 0 to t = 4. f. Calculate the total displacement of the particle from t = 0 to t = 4. g. Find all values of t in the interval 0 < t < 2 for which the speed of the particle is 1. h. Calculate the average velocity of the particle from t = 1 to t = 3.

3) Differential Equation Super FRQ (Calculator Inactive)

(a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. Draw a solution curve through the point (0,1).

(b) Describe all points where the particular solution to the differential equation y = f(x) has a horizontal tangent. (c) Verify that d²y/dx² = y³(1 + 3x²y²). Does y = f(x) have a point of inflection at the point (1,2)? Explain. (d) Does y=f(x) have a relative minimum, relative maximum, or neither at (1,2)? Explain.

(e) Write the equation of the line tangent to y = f(x) at x =1. Use this tangent to approximate the value of f(1.5). (f) For the approximation found in (d), determine whether it was an underestimate or overestimate of the actual value of f(1.5). (g) Find the solution y = f(x) to the given differential equation with the initial condition f(1) = 2. Use your solution to find f(0).

4) Name______________

Deriva�ve Graph Super FRQ (Calculator Inactive)

The func�on f is differen�able on the closed interval [-6,5] and sa�sfies f(2) = 3. The graph of f’, the deriva�ve of f, consists of a semicircle and three line segments, as shown in the figure above. a. Find and . f ( )− 6 f (5) b. Write an expression for f(x) that involves an integral. c. Find , , and f (4) f ′ (4) " f (4) d. Find all values of x where is horizontally tangent. Jus�fy. Determine whether a rela�ve f

minimum, rela�ve maximum, or neither is occurring at those values. Jus�fy your answers. e. Find the absolute maximum and absolute minimum values of and the x -values they f

occur at on the closed interval [-6,5]. f. On what intervals is increasing? On what intervals is decreasing? Jus�fy your answer. f f g. On what intervals is concave up? On what intervals is concave down? Jus�fy your f f

answer. h. On what intervals is decreasing and concave up? Jus�fy your answer. f i. For what value(s) of x does have a point of inflec�on? Jus�fy your answer. f j. Find , , and . Find each or explain why it does not exist. f ′′ ( )− 5 f ′′ (0) (2) f ′′ k. Let . Find the equa�on of the line tangent to the graph of g at x = 2. (x) (x) g = x * f l. Let . Find all cri�cal values of h(x). (x) (x) x h = f − 2 m. Find the average rate of change of f’ on the interval [-6, 5]. There is no point c, on [-6, 5] for

which f’(c) is equal to that average rate of change. Explain why this statement does not contradict the Mean Value Theorem.

n. Let . Find the slope of the line tangent to the graph of p at the point where (x) (x ) p = f 2 − 1 x = 2.

5) Name____________ Integral Defined Func�on Super FRQ (Calculator Inactive)

Graph of f

The figure above shows the graph of the piecewise func�on (composed of 2 semi-circles and two line segments) . For , the func�on is defined by f ≤ x ≤ 5 − 5 g

dtg (x) = ∫x

1f (t)

a. Find and . g (5) g ( )− 5 b. Find , , and g (4) g′ (4) " g (4) c. Find all values of x where has a horizontal tangent. Jus�fy. Determine whether a rela�ve g minimum, rela�ve maximum, or neither is occurring at those values. Jus�fy your answers. d. Find the absolute maximum and absolute minimum values of , and the x-values they g occur at, on the closed interval [-5,5]. e. On what intervals is increasing? On what intervals is decreasing? Jus�fy your answer. g g f. On what intervals is concave up? On what intervals is concave down? Jus�fy your g g answer. g. On what intervals is decreasing and concave up? Jus�fy your answer. g h. For what value(s) of x does g(x) have a point of inflec�on? Jus�fy your answer. i. Find , , and . Find each or explain why it does not exist. g′′ ( )− 4 g′′ (2) (3) g′′ j. Let , find . h (x) = g (2x )− 4 h′ (4) k. Let , write the equa�on of the tangent line to at . k (x) = 5x

g(x) g x = 4 l. valuate E lim

x→1 x 7x+62 −2g(2x) 2−

m. Let . Find m(1). Find m’(x) and evaluate m’(1). x dtm (x) = 3 + ∫x

3−f (t)

6) Integral Defined Func�on Super FRQ

The figure above shows the graph of the piecewise func�on (composed of semi-circles and six line segments) . For , the func�on is defined by f 0≤ x ≤10 − 1 g

dtg (x) = ∫x

4f (t)

a. Find and . g (9) g ( )− 9 b. Find , , and g ( )− 6 g′ ( )− 6 g′′ ( )− 6 c. Find all values of x where has a horizontal tangent. Jus�fy. Determine whether a rela�ve g minimum, rela�ve maximum, or neither is occurring at those values. Jus�fy your answers. d. Find the absolute maximum and absolute minimum values of on the closed interval g [-10,10]. e. On what intervals is increasing? On what intervals is decreasing? Jus�fy your answer. g g

The figure above shows the graph of the piecewise func�on (composed of semi-circles and six line segments) . For , the func�on is defined by f 0≤ x ≤10 − 1 g

dtg (x) = ∫x

4f (t)

f. On what intervals is concave up? On what intervals is concave down? Jus�fy your g g answer. g. On what intervals is increasing and concave down? Jus�fy your answer. g h. Find , , and . Find each or explain why it does not exist. g′′ (5) g′′ (4) g′′ ( )− 3 i. Let , find . h (x) = g (x )2 + 1 h′ (1) j. Let , write the equa�on of the tangent line to at . k (x) = 5x

g(x) g x = 8

. F ind k limx→2

g(3x) 2−x 7x+102−

7)

Area Volume Review Ques�on (Non-Calc Ac�ve)

Let R be the region bounded by the func�ons f (x) = and g(x) = x +1 as shown below. √x + 1

(a) Find the area of R.

(b) Find the volume of the solid generated when R is rotated around the x-axis.

(c) Write, but do not evaluate, an integral expression that gives the volume of the solid that is generated when R is rotated around the horizontal line y = 3.

(d) Write, but do not evaluate, an integral expression that gives the volume of the solid that is generated when R is rotated around the horizontal line y = −1.

(e) Write, but do not evaluate, an integral expression that gives the volume of the solid that is generated when R is rotated around the ver�cal line x = −2 .

(f) Write, but do not evaluate, an integral expression that gives the volume of the solid that is generated

when R is rotated around the ver�cal line x = 1.

(g) Region R is the base of a solid. For each x, where −1 ≤ x ≤ 0 , the cross sec�on of the solid taken

perpendicular to the x-axis is a rectangle whose height is 4 �mes the length of its base in region R. Write,

but do not evaluate, an integral expression that gives the volume of the solid. (How would this change if

it was an isosceles right triangle with a leg across the base?)

(h) Region R is the base of a solid. For each y, where 0 ≤ y ≤1 , the cross sec�on of the solid taken

perpendicular to the y-axis is a semicircle. Write, but do not evaluate, an integral expression that gives

the volume of the solid. (How would this change if it were an equilateral triangle? Square?)

(i) The ver�cal line x = k divides R into two regions with equal area. Write, but do not solve, an equa�on

involving integral expressions whose solu�on gives the value k.

(j) The horizontal line y = h divides R into two regions with equal area. Write, but do not solve, an

equa�on involving integral expressions whose solu�on gives the value h.

8)

Name______________ Table Super FRQ (Calculator Inac�ve)

t (seconds)

0 1 3 6 8 10

b’(t) (feet per second)

4.2 3.2 2.6 2.1 1.5 0.7

The volume of a spherical jumbo beach ball expands as air is being blown into it and maintains its spherical shape. The radius of the ball, in feet, is modeled by a twice-differen�able func�on b of �me t, where t is measured in seconds. For 0 < t < 10, the graph of b is concave down. The table above gives selected values of the rate of change b’(t) over the �me interval . The radius of the jumbo 00 ≤ t ≤ 1 beach ball is 5 feet when t = 1.

a) Using the table above, approximate b”(9). Interpret the meaning of b’’(9) in the context of the problem, using correct units.

b) Evaluate . Including units, explain the meaning of this in the context of the dt∫10

0b′′ (t)

problem. c) Es�mate the radius of the ball when t = 1.5 using the tangent line approxima�on at t = 1. Is your es�mate greater than or less than the true value? Give a reason for your answer. d) Use a right Riemann sum with five subintervals indicated by the data in the table to approximate

. Using correct units, explain the meaning of in terms of the radius of the ball. (t)dt∫10

0b′ (t)dt∫

10

0b′

e) Is your approxima�on in part d an overes�mate or an underes�mate? Explain your answer.

f) Find the rate at which the volume of the ball is changing with respect to �me at t = 1. Include units of measure.

g) Is there a �me t, for 0 < t <10 where b’(t) = 2? Jus�fy your answer.

h) Is there a �me t for 0 < t < 10, where b’’(t) = -0.3? Jus�fy your answer.

i) Use a Le� Riemann sum, with 5 subintervals as indicated by the table, to find the average value of b’(t) on the interval 0 < t < 10. Include units.

Name___________ Rate-in Rate-out Super FRQ (Calculator Active) 1) Coal is burning in a furnace, thus exhausting the resource. The rate at which coal is burning, measured in pounds per hour, is given by . At t = 2 hours a worker starts supplying additional coal (t) 8sin ( )B = t

2π into the furnace. The rate at which coal is being added, measured in pounds per hour, is given by

. The worker stops adding coal at t = 5 hours. At t = 0 there are 500 pounds of (t) 11 10cos( )S = + 254πt

coal in the furnace. (a) Find the total amount of coal, in pounds, added by the worker.

(t) dt 7.896 lbs∫5

2B = 2

(b) Is the amount of coal in the furnace increasing or decreasing at t = 4 hours? Explain.

therefore the amount of coal in the furnace is increasing (4) (4) .986 S − B = 1 > 0 at t = 4 hours. (c) Find S’(3). Explain, with units, the meaning of this in the context of the problem.

. The rate at which coal is being supplied is (3) .017 lbs per hour S′ = − 5 2 decreasing at a rate of 5.017 lbs per hour per hour at t = 3 hours. (d) Write a piecewise function A(t), that gives the total amount of coal in the furnace at any time

. 0 ≤ t ≤ 5

(t) 500 (t)dt if 0A = − ∫t

0B ≤ t ≤ 2

(t) 500 (t) dt (t)dt if 2A = + ∫t

2S − ∫

t

0B ≤ t ≤ 5

(e) Find the minimum and maximum amount of coal, in pounds, from time t = 0 to t = 5 hours and the times that they occur.

= a (t) (t) when t 4.37323 B = S =

(0) 500 A =

(2) 00 (t)dt 97.475A = 5 − ∫2

0B = 4

(4.37323) 00 (t) dt (t)dt 13.722A = 5 + ∫a

2S − ∫

a

0B = 5

(5) 00 (t) dt (t)dt 12.803A = 5 + ∫5

2S − ∫

5

0B = 5

Maximum amount of 513.722 lbs at t = 4.37323. Minimum amount of 497.475 lbs at t = 2. (f) Find the average rate of coal burned from t = 0 to t = 8 hours.

(t)dt .441 lbs hr81 ∫

8

0B = 4 /

(g) After t = 5 hours, the coal continues to burn until there are 0 pounds of coal remaining. Write an equation involving one or more integral expressions, that gives the time t = k, that the coal is exhausted.

(5) (t)dt 0A − ∫k

5B =

Name____________ Position Velocity Acceleration Super FRQ (Calculator Active) 2) A particle moves along the x-axis so that its velocity at time t is given by :

At time t = 0, the initial position of the particle is x = 5. a. Find the acceleration of the particle at time t = 5

(5) (5) .611 v′ = a = 7 b. Is the particle speeding up or slowing down at time t = 5.1? Justify your answer.

. The particle is speeding up at t = 5.1 (5.1) .652 and v (5.1) (5.1) .991 v = 7 ′ = a = 7 because v(5.1) and a(5.1) have the same sign. c. Find all times on the interval 0 < t < 5, that the particle is at rest. t = 0.958 and t = 3.734 because v(t) = 0 at those times. d. Write an expression involving an integral that gives the position x(t). Find the position of the particle at t = 2.

(t) (0) (t)dtx = x + ∫t

0v

(t) (t)dtx = 5 + ∫t

0v

(2) (t)dtx = 5 + ∫2

0v

(2) .603 x = 6

e. Calculate the total distance traveled by the particle from t = 0 to t = 4.

v(t)∣dt .761∫4

0∣ = 7

f. Calculate the total displacement of the particle from t = 0 to t = 4.

(t)dt .224∫4

0v = − 1

g. Find all values of t in the interval 0 < t < 2 for which the speed of the particle is 1. v(t)∣ ∣ = 1

t = 0.786 and t = 1.216 h. Calculate the average velocity of the particle from t = 1 to t = 3.

(t)dt .87021 ∫

3

1v = − 1

3) Differential Equation Super FRQ (Calculator Inactive)

(a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. Draw a solution curve through the point (0,1).

(b) Describe all points where the particular solution to the differential equation y = f(x) has a horizontal tangent.

when x = 0 or y = 0 dxdy = 0

(c) Verify that . Does y = f(x) have a point of inflection at the point (1,2)? xy (xy )dx2

d y2

= y3 + 3 2 3 Explain.

( ) (xy )ddx dx

dy = ddx

3

xydx2d y2

= y3 + 3 2dxdy

xy (xy )dx2d y2

= y3 + 3 2 3

(1 x y )dx2d y2

= y3 + 3 2 2

y = f(x) does not have a point of inflection at (1,2) because when = dx2d y2

/ 0 evaluated at (1,2) and does not change sign at that point.

(d) Does y=f(x) have a relative minimum, relative maximum, or neither at (1,2)? Explain.

Neither because evaluated at (1,2) is 8 which is not 0 therefore does dxdy

dxdy

not change sign at (1,2) (e) Write the equation of the line tangent to y = f(x) at x =1. Use this tangent to approximate the value of f(1.5).

(x ) y − 2 = 8 − 1

(1.5) (1.5 ) f ≈ 8 − 1 + 2

(1.5) f ≈ 6 (f) For the approximation found in (d), determine whether it was an underestimate or overestimate of the actual value of f(1.5).

Underestimate because evaluated in the first quadrant is positive dx2d y2

meaning y = f(x) is concave up. (g) Find the solution y = f(x) to the given differential equation with the initial condition f(1) = 2. Use your solution to find f(0).

ydxdy = x 3 C = − 8

5

dy dx 1y3 = x − 1

2y2 = x21 2

− 85

dy dx∫

1y3 = ∫

x y = √ 1

x +5 4− 2 /

− 12y2 = x2

1 2 + C (0) f = √ 15 4/

Name______________ Deriva�ve Graph Super FRQ (Calculator Inactive)

4)

The func�on f is differen�able on the closed interval [-6,5] and sa�sfies f(2) = 3. The graph of f’, the deriva�ve of f, consists of a semicircle and three line segments, as shown in the figure above. a. Find and . f ( )− 6 f (5)

( ) (x)dxf − 6 = 3 + ∫6−

2f ′

( ) π f − 6 = 2 − 1

(5) (x)dxf = 3 + ∫5

2f ′

(5) f = 6 b. Write an expression for f(x) that involves an integral.

(x) (t)dtf = 3 + ∫x

2f ′

c. Find , , and f (4) f ′ (4) " f (4)

(4) f = 3 + (x)dx∫4

2f ′ = 2

11 (4) f ′ = 1 (4) f ′′ = − 1

d. Find all values of x where has a horizontal tangent. Jus�fy. Determine whether a rela�ve f minimum, rela�ve maximum, or neither is occurring at those values. Jus�fy your answers. At x = -1 and x = 2 has a horizontal tangent. At x = -1 has a rela�ve maximum because f’ f f changes from posi�ve to nega�ve. At x = 2 has a rela�ve minimum because f’ changes from f nega�ve to posi�ve.

e. Find the absolute maximum and absolute minimum values of and the x -values they f occur at on the closed interval [-6,5].

( ) π f − 6 = 2 − 1

( ) π f − 1 = 2 + 3

(2) f = 3

(5) f = 6 Absolute max of which occurs at x = -1. Absolute min of 3 which occurs at x = 2. π 2 + 3 f. On what intervals is increasing? On what intervals is decreasing? Jus�fy your answer. f f f is increasing on [-6, -2] and [2, 5] because f’ is posi�ve. f is decreasing on [-2,2] because f’ is nega�ve. g. On what intervals is concave up? On what intervals is concave down? Jus�fy your answer. f f f is concave up on (0,3) because f’ is increasing. f is concave down on (-6, 0) and (3,5) because f’ is decreasing. h. On what intervals is decreasing and concave up? Jus�fy your answer. f f is decreasing and concave up on (0,2) because f’ is nega�ve and increasing. i. For what value(s) of x does have a point of inflec�on? Jus�fy your answer. f f has a point of inflec�on at x = 0 and x = 3 because f’ changes from increasing to decreasing or vice-versa.

j. Find , , and . Find each or explain why it does not exist. f ′′ ( )− 5 f ′′ (0) (2) f ′′

( ) 2 f ′′ − 5 = − 1/ = 0 f ′′ (0) does not exist because f ’ has a corner at x = 2. (2) f ′′

k. Let . Find the equa�on of the line tangent to the graph of g at x = 2. (x) (x) g = x * f

(2) g = 6 (x) (x) (x) g′ = f + x * f ′ (x) (2) (2) g′ = f + 2 * f ′

3 (2) g′ = angent line y (x ) T : − 6 = 3 − 2

l. Let . Find all cri�cal values of h(x). (x) (x) x h = f − 2

(x) (x) h′ = f ′ − 2 (x) 0 = f ′ − 2

(x) f ′ = 2 x = -6 and x = 3. m. Find the average rate of change of f’ on the interval [-6, 5]. There is no point c, on [-6, 5] for which f’(c) is equal to that average rate of change. Explain why this statement does not contradict the Mean Value Theorem.

5 ( 6)− −f (5) f ( 6)′ − ′ − = − 2

11 This does not contradict MVT because f’(x) is not differen�able on the en�re interval (-6,5). n.Let . Find the slope of the line tangent to the graph of p at the point where (x) (x ) p = f 2 − 1 x = 2.

(x) x (x ) p′ = 2 * f ′ 2 − 1

(2) (3) p′ = 4 * f ′

(2) p′ = 8

5) Name____________ Integral Defined Func�on Super FRQ (Calculator Inactive)

Graph of f

The figure above shows the graph of the piecewise func�on (composed of 2 semi-circles and two line segments) . For , the func�on is defined by f ≤ x ≤ 5 − 5 g

dtg (x) = ∫x

1f (t)

a. Find and . g (5) g ( )− 5

(5) dxg = ∫5

1f (x) = 2

5

( ) dxg − 5 = ∫5−

1f (x) = 2

3π

b. Find , , and g (4) g′ (4) " g (4)

(4) dxg = ∫4

1f (x) = 3

(4) 0 g′ =

(4) g′′ = − 1 c. Find all values of x where has a horizontal tangent. Jus�fy. Determine whether a rela�ve g minimum, rela�ve maximum, or neither is occurring at those values. Jus�fy your answers. g has a horizontal tangent at x = -3, x = 1, and x = 4 At x = -3 and x = 4, g has a rela�ve maximum because g’ changes from posi�ve to nega�ve At x = 1, g has a rela�ve minimum because g’ changes from nega�ve to posi�ve

d. Find the absolute maximum and absolute minimum values of , and the x-values they g occur at, on the closed interval [-5,5].

( )g − 5 = 23π

( ) π g − 3 = 2

(1) g = 0

(4) g = 3

(5) g = 2

5 g has an absolute maximum of at x = -3 and an absolute minimum of 0 at x = 1. π2 e. On what intervals is increasing? On what intervals is decreasing? Jus�fy your answer. g g g is increasing on [-5, -3] and [1, 4] because g’ is posi�ve. g is decreasing on [-3,1] and [4,5] because g’ is nega�ve. f. On what intervals is concave up? On what intervals is concave down? Jus�fy your g g answer. g is concave up on (-5,4) and (-1,2) because g’ is increasing. g is concave down on (-4, -1) and (2,5) because g’ is decreasing. g. On what intervals is decreasing and concave up? Jus�fy your answer. g g is decreasing and concave up on (-1,1) because g’ is nega�ve and increasing. h. For what value(s) of x does g(x) have a point of inflec�on? Jus�fy your answer. g has a point of inflec�on at x = -4, x = -1, and x = 2 because g’ changes from increasing to decreasing or vice-versa.

i. Find , , and . Find each or explain why it does not exist. g′′ ( )− 4 g′′ (2) (3) g′′

g′′ ( )− 4 = 0 oes not exists because g is not dif ferentiable at x g′′ (2) = d ′ = 2

(3) g′′ = − 1 j. Let , find . h (x) = g (2x )− 4 h′ (4)

(x) g (2x ) h′ = 2 ′ − 4 (4) g (4) h′ = 2 ′ (4) h′ = 0

k. Let , write the equa�on of the tangent line to at . k (x) = 5x

g(x) g x = 4

(4)k = 5(4)g(4) = 3

20

(x)k′ =(5x)2

5xg (x) g(x) 5′ − *

(4) k′ = 400

20 0 3 5* − * = − 380

Tangent line: (x ) y − 3

20 = − 380 − 4

l. valuate E lim

x→1 x 7x+62 −2g(2x) 2−

Therefore L’Hopital’s Rule applies g(2x) and xlim

x→12 − 2 = 0 lim

x→1x2 − 7 + 6 = 0

limx→1 x 7x+62 −

2g(2x) 2− = limx→1 2x 7−

4g (2x)′ = 5−4g (2)′ = − 5

8

m. Let . Find m(1). Find m’(x) and evaluate m’(1). x dtm (x) = 3 + ∫x

3−f (t)

(1) (x) dx πm = 3 + ∫1

3−f = 3 − 2

(x) (x) m′ = 3 + f

(1) (1) m′ = 3 + f = 3 + 0 = 3

6) Integral Defined Func�on Super FRQ

The figure above shows the graph of the piecewise func�on (composed of semi-circles and six line segments) . For , the func�on is defined by f 0≤ x ≤10 − 1 g

dtg (x) = ∫x

4f (t)

a. Find and . g (9) g ( )− 9

(9) (x)dxg = ∫9

4f = 2

7

( ) (x)dx 4 πg − 9 = ∫9−

4f = − 2 − 2

b. Find , , and g ( )− 6 g′ ( )− 6 g′′ ( )− 6

6 π g ( )− 6 = − 3 − 2 g′ ( )− 6 = 0 g′′ ( )− 6 = 4

c. Find all values of x where has a horizontal tangent. Jus�fy. Determine whether a rela�ve g minimum, rela�ve maximum, or neither is occurring at those values. Jus�fy your answers. g has a horizontal tangent at x = -6, x = 0, x = 4, and x = 8 At x = 8, g has a rela�ve maximum because g’ changes from posi�ve to nega�ve. At x = -6, g has a rela�ve minimum because g’ changes from nega�ve to posi�ve. At x = 0 and x = 4, neither occurs because g’ does not change signs.

d. Find the absolute maximum and absolute minimum values of on the closed interval g [-10,10].

( 0) 8 π g − 1 = − 2 − 2

( ) 6 π g − 6 = − 3 − 2

(8) g = 4

(10) g = 2 g has an absolute maximum of at x = 8 and an absolute minimum of 4 6 π− 3 − 2 at x = -6. e. On what intervals is increasing? On what intervals is decreasing? Jus�fy your answer. g g g is increasing on [-6, 8] because g’ is posi�ve. g is decreasing on [-10,-6] and [8,10] because g’ is nega�ve. f. On what intervals is concave up? On what intervals is concave down? Jus�fy your g g answer. g is concave up on (-7,-4) and (0,2) and (4,6) because g’ is increasing. g is concave down on (-1, 0) and (2,4) and (6,10) because g’ is decreasing. g. On what intervals is increasing and concave down? Jus�fy your answer. g g is increasing and concave down on (-1,0) and (2,4) and (6,8) because g’ is posi�ve and decreasing. h. Find , , and . Find each or explain why it does not exist. g′′ (5) g′′ (4) g′′ ( )− 3

g′′ (5) = 1 oes not exists because g is not dif ferentiable at x g′′ (4) = d ′ = 4

( ) 0 g′′ − 3 =

i. Let , find . h (x) = g (x )2 + 1 h′ (1)

(x) xg (x ) h′ = 2 ′ 2 + 1 (1) g (2) h′ = 2 ′ (1) h′ = 4

j. Let , write the equa�on of the tangent line to at . k (x) = 5x

g(x) g x = 8

(4)k = 5(8)g(8) = 4

40 = 110

(x)k′ =

(5x)25xg (x) g(x) 5′ − *

(8) k′ = 1600

40 0 4 15* − * = − 380

Tangent line: (x ) y − 1

10 = − 380 − 8

. F ind k limx→2

g(3x) 2−x 7x+102−

Therefore L’Hopital’s Rule applies (3x) and x 0lim

x→2g − 2 = 0 lim

x→2x2 − 7 + 1 = 0

limx→2

g(3x) 2−x 7x+102− = lim

x→2 2x 7−3g (3x)′ = 3−

3g (6)′ = − 36 = − 2

7)

Area Volume Review Ques�on (Non-Calc Ac�ve)

Let R be the region bounded by the func�ons f (x) = and g(x) = x +1 as shown below. √x + 1

(a) Find the area of R.

∫0

1−) x )dx (x ) x ∣ (√x + 1 − ( + 1 = 3

2 + 1 2 3/ − 21 2 − x 0

1− = 61

(b) Find the volume of the solid generated when R is rotated around the x-axis.

( ) x ) dxπ ∫0

1−√x + 1 2 − ( + 1 2 = 6

π

(c) Write, but do not evaluate, an integral expression that gives the volume of the solid that is generated when R is rotated around the horizontal line y = 3.

(3 (x)) 3 (x)) dxπ ∫0

1−− g 2 − ( − f 2

(d) Write, but do not evaluate, an integral expression that gives the volume of the solid that is generated when R is rotated around the horizontal line y = −1.

(f (x) ) g(x) ) dxπ ∫0

1−+ 1 2 − ( + 1 2

(e) Write, but do not evaluate, an integral expression that gives the volume of the solid that is generated when R is rotated around the ver�cal line x = −2 .

(y) f = y2 − 1

(y) g = y − 1

(g(y) ) f (y) ) dyπ ∫1

0+ 2 2 − ( + 2 2

(f) Write, but do not evaluate, an integral expression that gives the volume of the solid that is generated

when R is rotated around the ver�cal line x = 1.

(1 (y)) 1 (y)) dyπ ∫1

0− f 2 − ( − g 2

(g) Region R is the base of a solid. For each x, where −1 ≤ x ≤ 0 , the cross sec�on of the solid taken

perpendicular to the x-axis is a rectangle whose height is 4 �mes the length of its base in region R. Write,

but do not evaluate, an integral expression that gives the volume of the solid. (How would this change if

it was an isosceles right triangle with a leg across the base?)

ectangles V olume 4 (f (x) (x)) dxR : ∫0

1−− g 2

sosceles Right T riangle V olume (f (x) (x)) dxI : 21 ∫

0

1−− g 2

(h) Region R is the base of a solid. For each y, where 0 ≤ y ≤1 , the cross sec�on of the solid taken

perpendicular to the y-axis is a semicircle. Write, but do not evaluate, an integral expression that gives

the volume of the solid. (How would this change if it were an equilateral triangle? Square?)

emicircles V olume (g(y) (y)) dyS : 8π ∫

1

0− f 2

quilateral T riangle V olume (g(y) (y)) dyE : 4√3 ∫

1

0− f 2

quares V olume (g(y) (y)) dyS : ∫1

0− f 2

(i) The ver�cal line x = k divides R into two regions with equal area. Write, but do not solve, an equa�on

involving integral expressions whose solu�on gives the value k.

(x) (x) dxπ12 = ∫

k

1−f − g

(j) The horizontal line y = h divides R into two regions with equal area. Write, but do not solve, an

equa�on involving integral expressions whose solu�on gives the value h.

(y) (y) dyπ12 = ∫

k

0g − f

Name______________ Table Super FRQ (Calculator Inac�ve)

8)

t (seconds)

0 1 3 6 8 10

b’(t) (feet per second)

4.2 3.2 2.6 2.1 1.5 0.7

The volume of a spherical jumbo beach ball expands as air is being blown into it and maintains its spherical shape. The radius of the ball, in feet, is modeled by a twice-differen�able func�on b of �me t, where t is measured in seconds. For 0 < t < 10, the graph of b is concave down. The table above gives selected values of the rate of change b’(t) over the �me interval . The radius of the jumbo 00 ≤ t ≤ 1 beach ball is 5 feet when t = 1.

a) Using the table above, approximate b”(9). Interpret the meaning of b’’(9) in the context of the problem, using correct units.

(9) .4 feet per second per second b′′ ≈ 10 8−b (10) b (8)′ − ′ = − 0

b’’(9) means the rate at which the beach ball is being inflated is decreasing at a rate of 0.4 feet per second per second at t = 9 seconds.

b) Evaluate . Including units, explain the meaning of this in the context of the dt∫10

0b′′ (t)

problem.

dt (10) (0) .5∫10

0b′′ (t) = b′ − b′ = − 3

This means the rate at which the beach ball was being in�lated decreased by 3.5 feet per second from t =0 until t = 10 seconds. c) Es�mate the radius of the ball when t = 1.5 using the tangent line approxima�on at t = 1. Is your es�mate greater than or less than the true value? Give a reason for your answer.

.2(x ) y − 5 = 4 − 1

(1.5) .2(1.5 ) .1 feet b ≈ 4 − 1 + 5 = 7 Greater than true value because b(t) is concave down.

d) Use a right Riemann sum with five subintervals indicated by the data in the table to approximate

. Using correct units, explain the meaning of in terms of the radius of the ball. (t)dt∫10

0b′ (t)dt∫

10

0b′

(t)dt 1)(3.2) 2)(2.6) 3)(2.1) 2)(1.5) 2)(0.7) 9.1∫10

0b′ ≈ ( + ( + ( + ( + ( = 1

This means the ball’s radius increased by 19.1 feet from t = 0 to t = 10 seconds.

e) Is your approxima�on in part d an overes�mate or an underes�mate? Explain your answer.

Overes�mate because b’(t) is a decreasing func�on.

f) Find the rate at which the volume of the ball is changing with respect to �me at t = 1. Include units of measure.

πr V = 34 3

πr dtdV = 4 2

dtdr

π(5) b (1) dtdV = 4 2 ′

20π cubic feet per seconddtdV = 3

g) Is there a �me t, for 0 < t <10 where b’(t) = 2? Jus�fy your answer.

Since b’(t) is differen�able, b’(t) is con�nuous therefore IVT applies. Therefore, there is a �me t, for

0 < t < 10 where b’(t) = 2 because b’(0) > 2 > b’(10).

h) Is there a �me t for 0 < t < 10, where b’’(t) = -0.3? Jus�fy your answer.

Since b’(t) is differen�able, b’(t) is con�nuous therefore MVT applies. Therefore, there is a �me t, for

6 < t < 8 where b’’(t) = -0.3 because . .38 6−b (8) b (6)′ − ′ = − 0

i) Use a Le� Riemann sum, with 5 subintervals as indicated by the table, to find the average value of b’(t) on the interval 0 < t < 10. Include units.

(t)dt [(1)(4.2) 2)(3.2) 3)(2.6) 2)(2.1) (1.5)] .56 feet per second.110 ∫

10

0b′ ≈ 1

10 + ( + ( + ( + 2 = 2