AP Calculus AB : 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema- Exam Style questions with Answer- FRQ

Question:

The continuous function f is defined on the interval -4 ≤  x ≤ 3. The graph of f consists of two quarter circles and one line segment, as shown in the figure above. Let \(g(x)=2x+\int_0^x f(t)dt\).

(a) Find g(-3) . Find g'(x) and evaluate g'(-3) .
(b) Determine the x-coordinate of the point at which g has an absolute maximum on the interval -4 ≤  x ≤ 3. Justify your answer.
(c) Find all values of x on the interval -4 ≤  x ≤ 3 for which the graph of g has a point of inflection. Give a reason for your answer.
(d) Find the average rate of change of f on the interval-4 ≤  x ≤ 3.  There is no point c, -4 < c < 3,  for which f'(c) is equal to that average rate of change. Explain why this statement does not contradict the Mean Value Theorem.

▶️Answer/Explanation

Ans:

(a)

\(g(-3)=2(-3)+\int_{0}^{-3}f(t)dt = -6 -\frac{9\pi }{4}\)

\(g'(x)=\frac{d}{dx}\left ( 2x + \int_{0}^{x}f(t)dt \right )=2+f(x)\)

g'(-3) = 2 + f(-3) = 2 + 0 = 2

(b)

g'(x) = 0                        2 + f(x) = 0 

g'(x)                              f(x) = -2

                                       x = 5/2

                                    \(g(-4)=-8+\int_{0}^{-4}f(t)dt\)

                                    = -8 – 2π

\(g(5/2)= 5+\int^{\frac{5}{2}}_{0} f(t)dt = + \frac{5}{4}\)

\(g(3)= 6+\int_{0}^{3}f(t)dt = 6\)

x = 5/2 , because g’   going from t to – proves it as the only relative maximum and g(5/2) is greater than g at either endpoint.   

(c)

f'(x)                          g”(x) = d/dx    (g(x)) = f'(x)

The only point of inflection for g is at x = 0, since f'(x), which is equivalent to g”, only changes signs at x = 0 on the interval -4 ≤ x ≤ 3

(d)

Avg. Rate of change  = \(\frac{f(3)-f(-4)}{3-4}\)

\(\frac{-3-1}{3+4}=\frac{-2}{7}\)

Because mean value Theorem only applies when the function is continuous And differentiable on the interval, which doesn’t apply here since f(x) isn’t differentiable at x = 0.

Question:

Let f be the continuous function defined on [-4, 3] whose graph, consisting of three line segments and a semicircle centered at the origin, is given above. Let g be the function given by \(g(x)=\int_1^x f(t) dt\).

(a) Find the values of g(2) and g(-2).
(b) For each of g'(-3) and g”(-3) , find the value or state that it does not exist.
(c) Find the x-coordinate of each point at which the graph of g has a horizontal tangent line. For each of these points, determine whether g has a relative minimum, relative maximum, or neither a minimum nor a maximum at the point. Justify your answers.
(d) For -4 < x < 3, find all values of x for which the graph of g has a point of inflection. Explain your reasoning.

▶️Answer/Explanation

Ans:

(a)

\(g(2)= \int_{1}^{2}f(t)dt\)

\(g(2)= \frac{-1}{2}(1)\left ( \frac{1}{2} \right )\)

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

\(g(-2)= \int_{1}^{-2}f(t)dt\)

\(g(-2)= \frac{1}{2}\pi(1)^{2}-\left ( \frac{1}{2}(1)(3) \right )\)

\(g(-2)= \frac{\pi -3}{2}\)

(b)

g'(x) = f(x)

g'(-3) = 2

g”(x) = f'(x)

g”(-3) = 1

(c)

g'(x) = f(x) = 0

x = -1 , 1

At x = -1 g has a relative maximum because g'(x) = f(x) changes from positive to negative at x = 1  g has neither because g'(x) = f(x) does not change sign 

(d) 

g has inflection points where g”(x) = f'(x) changes sign. This occurs at x = -2 , 0 , 1

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:

Consider the differential equation \(\frac{dy}{dx}=2x – y.\)

(a) On the axes provided, sketch a slope field for the given differential equation at the six points indicated.

(b) Find \(\frac{d^{2}y}{dx^{2}}\) in terms of x and y. Determine the concavity of all solution curves for the given differential equation in Quadrant II. Give a reason for your answer.
(c) Let y = f(x) be the particular solution to the differential equation with the initial condition f (2) = 3.
Does f have a relative minimum, a relative maximum, or neither at x = 2 ? Justify your answer.
(d) Find the values of the constants m and b for which y = mx + b is a solution to the differential equation. 

▶️Answer/Explanation

Ans:

(b)

\(\frac{dy}{dx}= 2x – y\)

\(\frac{d^{2}y}{dx^{2}}= 2-\frac{dy}{dx}=2-(2x-y)\)

\(\frac{d^{2}y}{dx^{2}}= 2-2x+y\)

In Quadrant II, x < 0 and y > 0, So \(\frac{d^{2}y}{dx^{2}}= 2-2x+y>0,\)

Thus all solution curves in Quadrant II are concave up.

(c)

\(\frac{dy}{dx}= 2x-y=2.2-3 =1\)

Neither, as \(\frac{dy}{dx}\) ≠ 0   at x = 2.

(d)

\(\frac{dy}{dx}= 2x-y\),  y = mx + b

\(\frac{dy}{dx}= m = 2x – y\)

m = 2x – (mx + b)

m = (2-m)x – b,  equate coefficients

2 – m = 0

m = 2

-b = m

b = -m = -2.

m = 2,  b = -2

Question:

People enter a line for an escalator at a rate modeled by the function r given by

\(r(t)=\left\{\begin{matrix}
44\left ( \frac{t}{100} \right )^{3} \left ( 1-\frac{t}{300} \right )^{7}& for 0\leq t\leq 300 & \\
0&for t>300, &
\end{matrix}\right.\)

where r (t) is measured in people per second and t is measured in seconds. As people get on the escalator, they exit the line at a constant rate of 0.7 person per second. There are 20 people in line at time t = 0.
(a) How many people enter the line for the escalator during the time interval 0 ≤ t ≤ 300 ?
(b) During the time interval 0 ≤ t ≤ 300, there are always people in line for the escalator. How many people are in line at time t = 300 ?
(c) For t > 300, what is the first time t that there are no people in line for the escalator?
(d) For 0 ≤ t ≤ 300, at what time t is the number of people in line a minimum? To the nearest whole number, find the number of people in line at this time. Justify your answer.

▶️Answer/Explanation

Ans:

(a)

\(\int_{0}^{300}44\left ( \frac{t}{100} \right )^{3}\left ( 1-\frac{t}{300} \right )^{7}dt = 270\)

(b)

.7(300) = 210

20 + 270 = 290                                       290 – 210 = 80

(c)

(t-300)(.7) – 80 = 0

.7t – 210 – 80 = 0

.7t = +290

t = 414.2865

(d)

p = total people 

\(\frac{dp}{dt}=r(t)-.7\)

0 = r(t) – .7

t = 166.575

t = 33.013

\(p(t)= \int_{0}^{6}r(x)-.7 dx + 20\)

minimum at time t = 33.013 s when 4 people are in line

Question:

Let f be the function defined by f(x) = ex cos x.
(a) Find the average rate of change of f on the interval 0 ≤ x ≤ π .
(b) What is the slope of the line tangent to the graph of f at \(x = \frac{3\pi }{2}?\)
(c) Find the absolute minimum value of f on the interval 0 ≤ x ≤ 2π. Justify your answer.
(d) Let g be a differentiable function such that \(g\left ( \frac{\pi }{2} \right )=0.\) The graph of g’, the derivative of g, is shown below. Find the value of \(\lim_{x\rightarrow \pi /2}\frac{f(x)}{g(x)}\) or state that it does not exist. Justify your answer. 

▶️Answer/Explanation

Ans:

(a)

\(\frac{\int_{0}^{r}f'(x)dx}{r}=\frac{f(r)-f(0)}{r}=\frac{e^{r}cosr-e^{0}cos 0}{r}\)

\(\frac{e^{r}(-1)-1(1)}{r}=\frac{-e^{r}-1}{r}\)

(b)

\(\frac{df}{dx}=e^{x}cos x + e^{x}(-sinx)\)

\(=e^{x}(cos x -sinx)\)

\(=e^{3r/2}(0-(-1))=e^{3r/2}\)

(c)

from part B

\(f'(x)=e^{x}(cos x – sin x)= 0 \rightarrow e^{x}=0\)      or cox = sin x                         \(x = \frac{\pi }{4},\frac{5\pi}{4}\)

Minima either have a derivative at 0, undefined, or are boundary points.

So possible values : x = 0,  x = 2π,  x = π/4      x = 5π/4

f(x) = 1                f(x) = e        \(f(x)=e^{\pi /4}\left ( \frac{\sqrt{2}}{2} \right )\)                 \(f(x)=e^{5\pi /4}\left ( -\frac{\sqrt{2}}{2} \right )\)

Since \(e^{5\pi /4}\left ( -\frac{\sqrt{2}}{2} \right )\) is the only negative value of f(x), the absolute minimum of f(x) on 0 ≤ x ≤ 2π is \(-\frac{\sqrt{2}}{2}e^{5\pi /4}\) .

(d)

Note \(\lim_{x\rightarrow \frac{\pi }{2}}(g(x))=0\) since g(x) is differentiable (and thus continuous), and \(g\left ( \frac{\pi }{2} \right )=0.\) Also \(f\left ( \frac{\pi }{2} \right )=e^{\pi /2}\left ( cos\left ( \frac{\pi }{2} \right ) \right )=e^{\pi /2}(0)=0,\) and f(x) is continuous, So \(\lim_{x\rightarrow \frac{\pi }{2}}f(x)=0.\) So by L’ Hopital’s rule,

\(\lim_{x\rightarrow \frac{\pi }{2}}\left ( \frac{f(x)}{g(x)} \right )=\lim_{x\rightarrow \frac{2 }{2}}\left ( \frac{f'(x)}{g'(x)} \right )=\frac{e^{\pi /2}\left ( cos\frac{\pi }{2}-sin\frac{\pi }{2} \right )}{2}=\frac{-e^{\pi /2}}{2}\)

Question:

Consider the function y = f (x) whose curve is given by the equation 2y2 − 6 = y sin x for y > 0.
(a) Show that \(\frac{dy}{dx}=\frac{ycos x}{4y-sin x}.\)
(b) Write an equation for the line tangent to the curve at the point (0, \(\sqrt{3}\) ).
(c) For 0 ≤ x ≤ π and y > 0, find the coordinates of the point where the line tangent to the curve is horizontal.
(d) Determine whether f has a relative minimum, a relative maximum, or neither at the point found in part (c). Justify your answer.

▶️Answer/Explanation

Ans:

(a)

2y2-6 = y sin x

4y y’ = y cos x + y’ sin x

4y y’ – y’ sin x = y cos x

y'(4y-sinx) = y cos x

\(y’ = \frac{ycos x}{4y-sinx}\)

\(\frac{dy}{dx} = \frac{ycos x}{4y-sinx}\)

(b)

\(\frac{dy}{dx}At (0,\sqrt{3})=\frac{\sqrt{3}cos(0)}{4\sqrt{3}-sin(0)}=\frac{\sqrt{3}(1)}{4\sqrt{3}-(0)}=\frac{\sqrt{3}}{4\sqrt{3}}=\frac{1}{4}\)

\(y-\sqrt{3}=\frac{1}{4}(x-0)\)

(c)

2y2 – 6 = y sin x

\(\frac{dy}{dx}=\frac{ycos x}{4y – sinx}\)       y ≠0 : -6 = 0 sinx

\(0=\frac{ycos x}{4y – sinx}\)

                                                                                                      \(x = \frac{\pi }{2}: 2y^{2}-6=y sin(\frac{\pi }{2})\)

                                                                                                        \(2y^{2}-6=y (x)\)

0 = y cos x                                                                                      2y2 – y-6 =0

y ≠ 0     x ≠ 0                                                                                 (2y + 3) (y – 2) = 0

                                       Since

0 = y cos x               cos (0) = 1                                                       y = 2                      2y + 3 = 0

x = π/2                             and                                                                                             2y = -3

                                         0 ≠ 1                                          

f(x) has a horizontal tangent at \(\left ( \frac{\pi }{2},2 \right )\)

(d)

Since \(\frac{d^{2}y}{dx^{2}}<0\)   at \(\left ( \frac{\pi }{2},2 \right )\), f has a relative maximum at that point.

\(\frac{dy}{dx}=\frac{y cos x}{4y-sin x}\)

\(\frac{d^{2}y}{dx^{2}} = \frac{(4y-sinx)\left [ y(-sinx)+y’cos x \right ]-(y cos x)(4y’-cosx)}{(4y-sin x)^{2}}\)

At \(\left ( \frac{\pi }{2},2 \right )\) : \(\frac{d^{2}y}{dx^{2}} = \frac{(4.2-1)\left [ 2(-1)+0 cos x \right ]-(2(0))(4(0)-0)}{(4(2)-1)^{2}}\)

\(\frac{d^{2}y}{dx^{2}} = \frac{(8-1)(-2)}{(8-1)^{2}}=\frac{7(-2)}{7}=\frac{-2}{7}\)

                                                                                                            concave down

Question:

A company designs spinning toys using the family of functions \(y = cx\sqrt{4-x^{2}},\) where c is a positive constant. The figure above shows the region in the first quadrant bounded by the x-axis and the graph of \(y = cx\sqrt{4-x^{2}},\) for some c. Each spinning toy is in the shape of the solid generated when such a region is revolved about the x-axis. Both x and y are measured in inches.
(a) Find the area of the region in the first quadrant bounded by the x-axis and the graph of \(y = cx\sqrt{4-x^{2}},\) for c = 6.
(b) It is known that, for \(y = cx\sqrt{4-x^{2}},\frac{dy}{dx}=\frac{c\left ( 4-2x^{2} \right )}{\sqrt{4-x^{2}}}.\) For a particular spinning toy, the radius of the largest cross-sectional circular slice is 1.2 inches. What is the value of c for this spinning toy?
(c) For another spinning toy, the volume is 2π cubic inches. What is the value of c for this spinning toy? 

▶️Answer/Explanation

Ans:

(a)

\(y = 6\times \sqrt{4-x^{2}}=0\)

x = 0,     x = 2

\(A = \int_{0}^{2}6\times \sqrt{4-x^{2}}dx\)                                              u = 4-x2

                                                                                                                                    du = -2xdx

\(A = \int_{4}^{0}-3 \sqrt{u}du=3\int_{0}^{4}u^{\frac{1}{2}}du=3 u^{3/2}\cdot \frac{2}{3}|_{0}^{4}\)

A = 2(43/2 – 03/2) = 2(22.3/2 – 0) = 2(8) = 16

A = 16

(b)

Largest cross section where y is greatest (maximum of y on graph). 

Find max:

\(\frac{dy}{dx}=\frac{c(4-2x^{2})}{\sqrt{4-x^{2}}}=0 \rightarrow c(4-2x^{2})=0\)

                                                                                                                             4 = 2x2

                                                                                                                            \(x = \sqrt{2}\)

At \(x = \sqrt{2}\), y = 1.2 (largest radius of cross-section equals 1.2, which is max y value)

\(y = cx\sqrt{4-x^{2}}\)

\(1.2 = c\sqrt{2}\left ( \sqrt{4-(\sqrt{2})^{2}} \right )=c\sqrt{2}\left ( \sqrt{4-2} \right )=c\sqrt{2}(\sqrt{2})=2c\)

c = 1.2/2 = 0.6      →  c = 0.6

(c)

\(v = \pi \int_{0}^{2}y^{2}dx=\pi \int_{0}^{2}(cx\sqrt{4-x^{2}})^{2}dx = \pi c^{2}\int_{0}^{2}x^{2}(4-x^{2})dx=\pi c^{2}\int_{0}^{2}(4x^{2}-x^{4})dx\)

\(v = \pi c^{2}\left [ \frac{4x^{3}}{3}-\frac{x^{5}}{5}|_{0}^{2} \right ]= \pi c^{2}\left [ \left ( \frac{4(8)}{3}-\frac{32}{5} \right )-(0-0) \right ]\)

\(v = \pi c^{2}\left ( \frac{32(5)}{3(5)}-\frac{32(3)}{5(3)} \right )= \pi c^{2}\left ( \frac{2(32)}{15} \right )== \pi c^{2}\left ( \frac{64}{15} \right )\)

\(2\pi =\pi c^{2}\left ( \frac{64}{15} \right )\)

\(c^{2}=\frac{30}{64}\rightarrow c= \sqrt{\frac{30}{64}}=\frac{\sqrt{30}}{8}\)

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