Home / AP Calculus AB : 5.3 Determining Intervals on Which a Function Is  Increasing or Decreasing- Exam Style questions with Answer- FRQ

AP Calculus AB : 5.3 Determining Intervals on Which a Function Is  Increasing or Decreasing- Exam Style questions with Answer- FRQ

Question:

For 0 ≤ t ≤ 6, a particle is moving along the x-axis. The particle’s position, x (t), is not explicitly given.
The velocity of the particle is given by \(v(t)=2sin\left ( e^{t/4} \right )+1.\)  The acceleration of the particle is given by  \(a(t)=\frac{1}{2}e^{t/4}cos\left ( e^{t/4} \right )\) and x(0) = 2.
(a) Is the speed of the particle increasing or decreasing at time t = 5.5 ? Give a reason for your answer. 
(b) Find the average velocity of the particle for the time period 0 ≤ t ≤ 6.
(c) Find the total distance traveled by the particle from time t = 0 to t = 6.
(d) For 0 ≤ t ≤ 6,the particle changes direction exactly once. Find the position of the particle at that time.

▶️Answer/Explanation

Ans:

(a) The speed is increasing at t = 5.5 because v(5.5) < 0 and a(5.5) < 0

v (5.5) = -0.453
a (5.5) = -1.358

(b) 

\(\frac{1}{6-0}\int_{0}^{6}v(t)dt = 1.949\)

(c)

\(\int_{0}^{6}\left | v(t) \right |dt = 12.573\)

(d)

v(t) = 0 at t=5.1955223  and v(t) changes sign at t = 5.1955223

\(X(5.195223)= 2+\int_{0}^{5.195523}v(t)dt\)

Position at t = 5.195523 is 14.134

Question:

On a certain workday, the rate, in tons per hour, at which unprocessed gravel arrives at a gravel processing plant is modeled by \(G(t)=90+45 Cos\left ( \frac{t^{2}}{18} \right ),\)  where t is measured in hours and  0 ≤ t ≤ 8. At the beginning of the workday (t = 0), the plant has 500 tons of unprocessed gravel. During the hours of operation,0 ≤ t ≤ 8 the plant processes gravel at a constant rate of 100 tons per hour.

(a) Find G’(5). Using correct units, interpret your answer in the context of the problem.

(b) Find the total amount of unprocessed gravel that arrives at the plant during the hours of operation on this workday.

(c) Is the amount of unprocessed gravel at the plant increasing or decreasing at time t = 5 hours? Show the work that leads to your answer.

(d) What is the maximum amount of unprocessed gravel at the plant during the hours of operation on this workday? Justify your answer.

▶️Answer/Explanation

Ans:

(a)

\(G(t)=90+45 Cos\frac{t^{2}}{18}\)

\(G'(t)=-5t sin\left ( \frac{t^{2}}{18}  \right )\)

G’(5) = – 24.588 tons/ hr2

This means that the rate at which unprocessed gravel arrives at the processing plant is changing by -24.588 tons per hour per hour, or decreasing by 24.588 tons per hour per hour, at   t = 5 hours.

(b)

\(\int_{0}^{8}\left [ 90+45 cos\left ( \frac{t^{2}}{18} \right ) \right ]dt= 825.551 tons\)

(c)

Let V(t) be the amount of unprocessed gravel.

V’(t) is the rate of which the amount of unprocessed gravel is changing.

V’(t) = G(t) – 100

V’(5) = G(5) – 100

\(=90+45 cos \left ( \frac{52}{18} \right )-100\)

V’(5) = – 1.859

Since V’(5) is negative the amount of unprocessed gravel is decreasing at time t = 5 hours

(d)

V’(t) = 0   at    t = ?

0 = G(t)  – 100

\(100=90+45 cos \left ( \frac{t^{2}}{18} \right )\)

\(10=45 cos \left ( \frac{t^{2}}{18} \right )\)

T = 4.923

\(v(t)-v(0)=\int_{0}^{t}\left ( G(x)-100 \right )dx\)

\(v(t)=\int_{0}^{t}\left ( G(x)-100 \right )dx+v(0)\)

V (0) = 500

V(4.923) = 635.76

V(8) = 525.551

Since v(t) is on a closed interval [0, 8 ], the maximum amount must occur at an endpoint or at a critical value after evaluating the amount of unprocessed gravel at   t = 0, t = 4.923,  and t = 8, the amount of unprocessed gravel is highest at t = 4.923, with 635.376 tons of unprocessed gravel.

 Question:

A particle moves along a straight line. For 0 ≤ t ≤ 5, the velocity of the particle is given by \(v(t)= -2 + (t^{2}+3t)^{6/5} – t^{3}\), and the position of the particle is given by s(t). It is known that s(0) = 10.
(a) Find all values of t in the interval 0 ≤ t ≤ 4 for which the speed of the particle is 2.
(b) Write an expression involving an integral that gives the position s(t). Use this expression to find the position of the particle at time t = 5.
(c) Find all times t in the interval 0 ≤ t ≤ 5 at which the particle changes direction. Justify your answer.
(d) Is the speed of the particle increasing or decreasing at time t = 4 ? Give a reason for your answer.

▶️Answer/Explanation

Ans:

(a)

|v(t) | = 2                             2 ≤ t ≤ 4

t = 3.128,  3.473

(b)

\(s(t)= 10 + \int_{0}^{t}v(x)dx\)

\(s(5)= 10 + \int_{0}^{5}v(x)dx = -9.207\)

(c)

v(t) = 0

t = 0.536 ,  3.318

the particle changes direction at 0.536 because v(t) < 0 for (0, 0.536) and v(t) > 0 for (0.536, 3.318). The particle changes direction at 3.318 because v(t) > 0 for (0.536,  3.318) and v(t) < 0 for t> 3.318

(d)

v(4) < 0 

a(4) = v'(4) <0

The speed is increasing at t = 4 because both v(4) and a(4) = v'(4) are negative.

Question:

The figure above shows the graph of f’, the derivative of a twice-differentiable function f, on the closed interval 0 ≤ x ≤ 8.  The graph of f’ has horizontal tangent lines at x = 1, x = 3, and x = 5. The areas of the regions between the graph of f’ and the x-axis are labeled in the figure. The function f is defined for all real numbers and satisfies f (8) = 4.
(a) Find all values of x on the open interval 0 < x < 8 for which the function f has a local minimum. Justify your answer.
(b) Determine the absolute minimum value of f on the closed interval 0 ≤ x ≤ 8.  Justify your answer.
(c) On what open intervals contained in 0 < x < 8  is the graph of f both concave down and increasing? Explain your reasoning.
(d) The function g is defined by \(g(x)=(f(x))^{3}\), if \(f(3)= -\frac{5}{2},\) find the slope of the line tangent to the graph of g at x = 3.

▶️Answer/Explanation

Ans:

(a)

X = 6

f has a local minmum at x = 6, because the graph of f’ changes from negative to positive at x = 6, so using the first derivative test and f has a critical number at x = 6, f has a local minimum 

(b)

local minimum  = x = 6                                                                                                                                            f(8) = 4

\(f(6)\rightarrow \int_{6}^{8}f'(x)dx = 7 = f(8)-f(6)=4-f(6)\)                                                                      f(6) = -3

\(f(0)\rightarrow \int_{0}^{8}f'(x) = 12 = f(8)-f(0)=4-f(0)\)                                                                       f(0) = -8

The absolute minimum value of f on the interval 0 ≤ x ≤ 8  is -8 because it is the lowest value for the among the endpoints and critical numbers.

(c) \({f}” < 0\)

The open intervals where the graph of f is both concave down and increasing is (0, 1) u (3, 4), or 0 < x < 1 and 3 < x 4, because using the graph of f’, when the graph of f’ is positive and the slope of f’ is negative that means that f is increasing and f” is negative so f is both concave down and increasing.

(d)

g'(x) = 3 (f(x))2. f'(x)

g'(3) = 3 (f(3))2 . f'(3) = 3 (-5/2)2 . (4)

= 3 (25/4)4 = 75

g'(3) = 75

Question:

The function f is defined on the closed interval [-5, 4] . The graph of f consists of three line segments and is shown in the figure above. Let g be the function defined by \(g(x)=\int_{-3}^{x}f(t)dt.\)

(a) Find g(3) .
(b) On what open intervals contained in -5 < x < 4 is the graph of g both increasing and concave down? Give a reason for your answer.
(c) The function h is defined by \(h(x)= \frac{g(x)}{5x}\). Find h'(3).
(d) The function p is defined by p(x) = f(x2 – x). Find the slope of the line tangent to the graph of p at the point where x = -1.

▶️Answer/Explanation

Ans:

\(g(x)= \int_{-3}^{3}f(t)dt\)

\(= \frac{1}{2}(5)(4)-\frac{1}{2}(1)(2)\)

= 10 – 1 = 9

(b)

g'(x) > 0 ⇔ f(x) > 0

g”(x) < 0 ⇔ f'(x) < 0

(-5, -3),  (0, 2)

(c)

\(h(x)=\frac{g(x)}{5x}\)

\(h'(x)= \frac{(5x)(g'(x))-g(x).5}{25x^{2}}\)

\(h'(3)= \frac{(15)(f(3))-g(3).5}{9.25}=\frac{(15)(-2)-(9)(5)}{9.25}\)

(d)

p(x) = f(x2 – x)

p'(x) = f'(x2-x) . (2x-1)

p'(-1) = f'(1+1) . (-2-1) = f'(2).-3

                                                                 = (-2) (-3)

                                                                 = 6

\({f}'(2)=\frac{-4-4}{4-0}=\frac{-8}{4}=-2\)

Question:

The rate at which rainwater flows into a drainpipe is modeled by the function R, where \(R(t)=20sin\left ( \frac{t^{2}}{35} \right )\) cubic feet per hour, t is measured in hours, and 0 ≤ t ≤ 8. The pipe is partially blocked, allowing water to drain out the other end of the pipe at a rate modeled by D(t) = -0.04t3 + 0.4t2 + 0.96t cubic feet per hour, for 0 ≤ t ≤ 8. There are 30 cubic feet of water in the pipe at time t = 0.
(a) How many cubic feet of rainwater flow into the pipe during the 8-hour time interval 0 ≤ t ≤ 8?
(b) Is the amount of water in the pipe increasing or decreasing at time t = 3 hours? Give a reason for your answer.
(c) At what time t, 0 ≤ t ≤ 8, is the amount of water in the pipe at a minimum? Justify your answer.
(d) The pipe can hold 50 cubic feet of water before overflowing. For t > 8, water continues to flow into and out of the pipe at the given rates until the pipe begins to overflow. Write, but do not solve, an equation involving one or more integrals that gives the time w when the pipe will begin to overflow.

▶️Answer/Explanation

Ans:

(a)

\(\int_{0}^{8}R(t)dt = \int_{0}^{8}20 sin\frac{t^{2}}{35}dt=76.570ft^{3}\)

During the eight hour interval, about 76.570 cubic feet of water flow into the drainpipe

(b)

Total water : T(x)                       \(T'(x) = R(x) – F(x) = 20in\frac{t^{2}}{35}+.04t.^{3}-.4t^{2}-.96t\)

T'(3) = 20 in 9/35 + .04(27) – .4(0) – .96(3) : : 314 < 0

after three hours, the amount of water in the pipe is decreasing because the derivative of the amount of water (the difference between water interring and leaving) is less than zero at 3 hours.

(c)

T'(t) = 0     @    t = 0,  3.2716584

\(T(t)=T(0)+\int_{0}^{t}T'(t)dt=30+\int_{0}^{t}T'(t)dt\)

T(0) = 30

T(3.272) = 27.965

T(8) = 48.544

after testing all critical numbers and endpoints for their values, the amount of water in the pipe achieves a minimum value of about 27.965 after about 3.272 hours.

(d)

50 = T(w)

\(50 = 30+\int_{0}^{w}T'(t)dt\)

\(20 = \int_{0}^{w}T'(t)dt=\int_{0}^{w}\left [ R(t)-D(t) \right ]dt\)

Question:

For t ≥ 0, a particle moves along the x-axis. The velocity of the particle at time t is given by \(v(t)=1+2 sin\left ( \frac{t^{2}}{2} \right )\) The particle is at position x = 2 at time t = 4.
(a) At time t = 4, is the particle speeding up or slowing down?
(b) Find all times t in the interval 0 < t < 3 when the particle changes direction. Justify your answer.
(c) Find the position of the particle at time t = 0.
(d) Find the total distance the particle travels from time t = 0 to time t = 3. 

▶️Answer/Explanation

Ans:

(a)

\(v(4)=1+2 sin\left ( \frac{4^{2}}{2} \right )\)

v(4) = 2.979

\(v'(t)=2 cos\left ( \frac{t^{2}}{2} \right )(t)\)

\(v'(4)=2 cos\left ( \frac{4^{2}}{2} \right )(4)\)

v'(4) = -1.164

Showing down because v(A) is positive and v'(4) is negative.

(b)

\(1+2sin\left ( \frac{t^{2}}{2} \right )=0\)

t = 2.707

The particle changes direction one time at t = 2.707 because v(t) = 0 and v(t) changes from positive to negative. 

(c)

\(2+\int_{4}^{0}v(t)dt = -3.815\)

(d)

\(2+\int_{0}^{3}|v(t)|dt = 5.301\)

Question:

The figure above shows the graph of f’, the derivative of a twice-differentiable function f, on the interval [-3, 4] . The graph of f’ has horizontal tangents at x =- 1, x =1, and x = 3. The areas of the regions bounded by the x-axis and the graph of f’ on the intervals [-2, 1] and [1, 4] are 9 and 12, respectively.
(a) Find all x-coordinates at which f has a relative maximum. Give a reason for your answer.
(b) On what open intervals contained in -3 < x < 4 is the graph of f both concave down and decreasing? Give a reason for your answer.
(c) Find the x-coordinates of all points of inflection for the graph of f. Give a reason for your answer.
(d) Given that f(1) = 3,, write an expression for f(x) that involves an integral. Find f(4) and f(-2).

▶️Answer/Explanation

Ans:

(a) 

∴ f(x) has a relative maximum at x = -2 because f'(x) switches from positive to negative of this point.

(b) The graph of f is both concave down and decreasing on the intervals (-2, -1) and (1, 3) because on these intervals f'(x) is negative and also f”(x) is negative.

(c)

x = -1,  1, 3

The x – coordinates of the points of inflection for the graph off f are x = -1,  x = 1,  x = 3, This is because at these points, f”(x) switches signs.

(d)

\(f(x)=\int_{1}^{x}f'(t)dt+3\)

\(f(4)=\int_{1}^{(4)}f'(t)dt+3\)                                                    \(f(-2)=\int_{1}^{(-2)}f'(t)dt+3\)

=  (12) + 3                                                                                            = (9) + 3

f(4) = -9                                                                                         f(-2) = 12

Question:

The function f is differentiable on the closed interval [−6, 5 ] and satisfies f (−2) = 7. The graph of f’, the derivative of f, consists of a semicircle and three line segments, as shown in the figure above.
(a) Find the values of f (−6) and f (5).
(b) On what intervals is f increasing? Justify your answer.
(c) Find the absolute minimum value of f on the closed interval [−6, 5 ]. Justify your answer.
(d) For each of f”(-5) and f “(3 ), find the value or explain why it does not exist.

▶️Answer/Explanation

Ans:

(a)

\(f(-6)=\left ( \int_{-2}^{-6}f'(x)dx \right )+f(-2)\)

f(-6) = 3

\(f(5)=f(-2)+\int_{-2}^{5}f'(x)dx\)

f(5) = 10 – 2π

(b)

f is increasing on x = [-6, -2]

u [2, 5], since f’> 0 on the interval x ∈ [-6, -2]   v[2, 5]

(c)

The absolute minimum endpoints of f or [-6, 5] is              f(-6) = 3

7 – 2 π , since                                                                                f(5) = 10 – 2π

f(2)  < f (5) and f(-6)                                                                 critical points

the endpoints                                                                            f’ = 0

and f(2)  < f(-2) the                                                                 f(-2) = 7

other critical points,                                                               f(2) = 7.2 π

by EVT

(d)

\(f”(-5)=\frac{-1}{2}\)

f”(3)  = DNE, as the 

\(\lim_{x\rightarrow 3^{+}}\frac{f'(x)-2}{x-3}\neq \lim_{x\rightarrow 3^{-}}\frac{f'(x)-2}{x-3}\)

Therefore it is impossible to take a derivative at x = 3 in f’

Question

Fish enter a lake at a rate modeled by the function E given by \(E(t)=20 + 15 sin\left ( \frac{\pi t}{6} \right ).\) Fish leave the lake at a rate modeled by the function L given by \(L(t) = 4 +2^{0.1t^2}\) . Both E (t) and L (t) are measured in fish per
hour, and t is measured in hours since midnight (t = 0).
(a) How many fish enter the lake over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5) ? Give your answer to the nearest whole number.
(b) What is the average number of fish that leave the lake per hour over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5) ?
(c) At what time t, for 0 ≤ t ≤ 8, is the greatest number of fish in the lake? Justify your answer.
(d) Is the rate of change in the number of fish in the lake increasing or decreasing at 5 A.M. (t = 5) ? Explain your reasoning. 

▶️Answer/Explanation

Ans:

(a)

\(\int_{0}^{5}E(t)dt\approx 153 fish\)

(b)

\(\frac{1}{5}\int_{0}^{5}L(t)dt\approx \frac{30.295}{5}\approx 6.059\) fish per hour leave the lake

(c)

E(t) – L(t) = 0

t ≈ 6.204

E(t) – L(t) is the rate at which the number of fish is changing

At time t = 6.204, the greatest number of fish in the 8 hour period are in the lake. This is because E(t) – L(t) is positive from t = 0 + 0    t = 6.204 indicating that the number of fish in the lake is increasing over (0, 3.204), but E(t) – L(t) is negative from t = 6.204 to  t = 8, which means the number of fish are decreasing in this time period, so the number of fish in the lake is greatest at t = 6.204 hours

(d)

\(\frac{d}{dt}\left ( 16+15 sin(\frac{\pi t}{6})-2^{0.1t2} \right )|_{t=5}\approx -10.723 fish / hour^{2}\)

Since the derivative of E(t) – L(t) at t = 5 is negative, the rate of change in the number of fish in the lake is decreasing

Question

Let f be the function given by f(x)=(lnx)(sinx).The figure above shows the graph of f for \(0<x\leq 2\pi \) The function g is defined by g(x )=\(g(x)=\int_{1}^{x}f(t) dt ~for~ 0<x\leq2 \pi \).

(a) Find g(1) and g'(1)

(b) On what intervals, if any, is g increasing? Justify your answer.

(c) For \(0<x\leq 2\pi \),  find the value of x at which g has an absolute minimum. Justify your answer.

(d) For  \(0<x\leq 2\pi \) is there a value of x at which the graph of g is tangent to the x-axis? Explain why or why not.

▶️Answer/Explanation

(a)\(g(1)=\int_{1}^{1}f(t)dt=0\) and \(g'(1)=f(1)=0 \)

(b) Since g'(x) = f(x), g is increasing on the interval 1≤ x ≤ because \(f(x) > 0 for 1 < x < \pi.\)

(c) For\( 0 < x < 2\pi, g'(x) = f(x) = 0\) when \(x = 1, \pi.\) g’ f changes from negative to positive only at x = 1. The absolute minimum must occur at x = 1 or at the right endpoint. Since g(1) = 0 and \(g(2\pi)=\int_{1}^{2\pi}f(t)dt=\int_{1}^{\pi}f(t)dt+\int_{\pi}^{2\pi}f(t)<0\) by comparison of the two areas, the absolute minimum occurs at \(x = 2π.\)

(d) Yes, the graph of g is tangent to the x-axis at x = 1 since g(1) = 0 and g'(1) = 0.

Scroll to Top