MA 16500 Midterm III Study Guide

How many inflection points does the graph of the function y=f(x)=x↑(5) −5x↑(4) +25x have?

# of inflection points = 1

The first derivative of a function f is given by f′(x) = (x+2)↑(2)(x+1)(x−1)↑(3)(x−3)↑(2)(x−5). Find the values of x for which the function f takes a local maximum and local minimum.

Local Max at x = 1 Local Min at x = -1, 5

Find the absolute maximum/minimum and local maximum/minimum of the function defined by f(x) = 3x4 −16x3 +18x2 on the closed interval [−1, 4].

Local Max: f(1) = 5 at x = 1 Local Min: f(0) = 0 at x = 0 f(3) = -27 at x = 3 Absolute Max: f(-1) = 37 at x = -1 Absolute Min: f(3) = -27 at x = 3

What is the largest area of an isosceles triangle inscribed in a circle of radius 1?

Max possible area = (3 * √(3)) / 4

A farmer has 400 feet of fencing to build a rectangular pen to contain chickens. The pen needs to be in the shape of a rectangle with a straight divider in the middle that separates the pen into two congruent rectangles. What is the maximum possible area inside the pen? (see pg. 6 in the study guide for diagram)

Max possible area = 20000 / 3

A rectangular box has to be made with the width being twice as long as the length (with a bottom but WITHOUT a top). If the surface area of the box is 400 cm², what is the height of the box with the largest volume?

Max possible volume = (20/9)*√(6)

What is the maximum area of a rectangle whose base is on the x-axis, having the remaining two vertices on the graph of y = 9 − x², and lying above the x-axis?

Maximum Area = 12√(3)

A cone-shaped drinking cup is made from a circular piece of paper of fixed radius R by cutting out a sector and joining the edges CA and CB. Find the maximum capacity of such a cup. (see pg. 7 in the study guide for diagram)

Maximum Capacity = ((2√(3)) / 27)πR³

What is the smallest possible area of a triangle formed by the (x²/3²) + (y²/7²) = 1 coordinate axes and a line tangent to the ellipse in the first quadrant?

Smallest Area = 21

11.12 in the Study Guide

When x = (6 / √(5)) km, the time T is minimum.

(a) Find the critical numbers of the function f(x) = x↑(8)(x − 4)↑(7). (b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers? (c) What does the First Derivative Test tell you that the Second Derivative test does not?

a) Critical Numbers: 0, (32/15), 4 b) 2nd Derivative Test: x = 0 inconclusive x = (32/15) local min x = 4 inconclusive c) 1st Derivative Test x = 0 local max x = (32/15) local min x = 4 neither

Draw the graph of the following function: y = f(x) = 1 / (x² - 16)

ans key: pg 66

Draw the graph of the following function: y = f(x) = x / (x² - 16)

ans key: pg 68

Draw the graph of the following function: y = f(x) = x² / (x² - 16)

ans key: pg 70

Draw the graph of the following function: y = f(x) = x³ / (x² - 16)

ans key: pg 74

Draw the graph of the following function: y = f(x) = x / (x² + 16)

ans key: pg 76

Draw the graph of the following function: y = f(x) = x³ / (x² + 1)

ans key: pg 79

Draw the graph of the following function: y = f(x) = e↑(-x)sin(x) on [0, 2π]

ans key: pg 81

Draw the graph of the following function: y = f(x) = ln(x²-10x+24)

ans key: pg 84

Consider the function f(x) = x↑(4) − 2x↑(2) + 7x − 2 over the interval [−2, 2]. Does it satisfy the conditions for the Mean Value Theorem to hold? If it does, find the value(s) c ∈ (−2, 2) such that f′(c) = (f(2) − f(−2)) / (2−(−2)).

c = 0, ±1 (all of which are in (-2, 2))

Consider the function y = f (x) = x↑(2/3) over the interval [−1, 1]. Does it satisfy the conditions for the Mean Value Theorem to hold? Do we have any value c ∈ (−1, 1) such that f′(c) = ((f (1) − f (−1)) / (1−(−1)))

continuous on [-1, 1] NOT differentiable ∴ no value of c

lim (x → (π/2)⁻) (2x - π) * tan(x)

lim (x → (π/2)⁻) 2/(-csc²(x)) = -2

lim (x → (π/2)⁻) (5tan(x))↑(cos(x))

lim (x → (π/2)⁻) y = 1

L'Hospital's Rule lim (x → 0) (3x - sin(3x)) / (5x - tan(5x))

lim (x → 0) ((3 * 3) / (-5 * 2 * 5)) * (3/5) = -27/250

L'Hospital's Rule lim (x → 0) ((e↑(7x) - cos(2x))/(tan(3x))

lim (x → 0) ((7e↑(7x) - 2sin(2x)) / (3sec↑(2)(3x)) = 7/3

L'Hospital's Rule lim (x → 0) (ln(cos(5x))) / (x↑(2))

lim (x → 0) (-5/2) * 5 = -25/2

L'Hospital's Rule lim (x → 0) (ln(sin(x) / x)) / x↑(2)

lim (x → 0) (-sin(x) / x) / (4(sin(x) / x) + 2cos(x)) = -1/6

L'Hospital's Rule lim (x → 0) (tan(x) - x) / x↑(3)

lim (x → 0) (2sec↑(3)(x)sin(x)) / (6x) = 1/3

L'Hospital's Rule lim (x → 0) (1-cos(x)) / (3x↑(2))

lim (x → 0) (cos(x) / 6) = 1/6

L'Hospital's Rule lim (x → 0) (sin(x)) / (1 - x↑(2))

lim (x → 0) (sin(x)) / (1 - x↑(2)) = 0

lim (x → 0) x↑(2) * tan(1/(5x↑(2) + 2))

lim (x → 0) 0 * tan(1/2) = 0

lim (x → 0⁺) x * ln(2x)

lim (x → 0⁺) (-x) = 0

lim (x → 0⁺) x * ln(3 + (5/x))

lim (x → 0⁺) 5x / (3x + 5) = 0

lim (x → 0⁺) (1-5x)↑(1/x)

lim (x → 0⁺) y = e↑(-5)

lim (x → 1) ((x / (x-1)) - (1 / ln(x)

lim (x → 1) (ln(x) + 1) / (ln(x) + 2) = 1/2

lim (x → 4) (1/(√(x) - 2)) - (4/(x - 4))

lim (x → 4) (1/(2√x)) / 1 = 1/4

lim (x → ∞) √(x↑(2) - 5x + 7) - x

lim (x → ∞) (-5 + (7/x)) / (√(1-(5/x) + (7/x)) + 1) = -(5/2)

L'Hospital's Rule lim (x → ∞) (ln(x) / √x)

lim (x → ∞) (2 / √x) = 0

lim (x → ∞) 2x * tan(1/(3x))

lim (x → ∞) (2/3)sec↑(2)(1/(3x)) = 2/3

lim (x → ∞) x↑(2) * tan(1/(5x↑(2) + 2))

lim (x → ∞) sec↑(2)(1/(5x↑(2) + 2)) * (5x↑(4) / (5x↑(4) + 2)↑(2)) = 1/5

lim (x → ∞) (e↑(x) + x)↑(1/x)

lim (x → ∞) y = e

lim (x → ∞) ((x+3)/(x-2))↑(4x+1)

lim (x → ∞) y = e↑(20)

lim (x → ∞) (1 + (3/x))↑(7x)

lim (x → ∞) y = e↑(21)

The second derivative of a function f is given by f′′(x) = (x + 5)↑(3)(x + 2)↑(2)(x − 2)↑(5)(x − 3)↑(3)(x − 6)↑(2). Find the x-coordinates of all the inflection points.

x = -5, 2, 3

Suppose a rectangular area is to be surrounded by a concrete walkway that is 2 meter wide on the East and West and 5 meters wide on the North and South. If the area inside the walkway is to be 100 square meters, what should the interior width of the enclosed area be (labeled x in the diagram above) in order to minimize the amount of concrete used. (see pg. 8 in the study guide for diagram)

x = 2√(10)

Find the equation of a line passing through the point (3,2), which cuts off the least amount of area from the first quadrant.

y = (-2/3x) +4

Determine the exact value of tan⁻¹(7) - tan⁻¹(-1/7)

π/2

Determine the exact value of sin⁻¹(1/5) + cos⁻¹(1/5)

π/2

Find the estimate of sin(46°) using the linear approximation of the function f(x)=sin(x) at a = π/4.

≈ ((√2) / 2) + ((√2) / 360)

Find the formula for the linear approximation to f(x) = e↑(x) at x=a = 0 and use it to approximate e↑(-0.05).

≈ 0.95

Find the formula for the linear approximation to f(x) = √x at x=a = 1 and use it to approximate √1.1.

≈ 1.05

Find the estimate of ↑(3)√8.012 using the linear approximation of the function f(x)=↑(3)√x at x=8.

≈ 2

The slant height of a right circular cone is the distance from the edge of the base of the cone to the vertex of the cone. What is the maximum volume of a right circular cone with slant height 5 cm. (see pg. 8 in the study guide for diagram)

((250√(3)) / 27)π

Determine how the concavity changes for the function f(x) = (1/2)x − sin(x) on the interval (0, 3π).

(0, π) - Concave Up (π, 2π) - Concave Down (2π, 3π) - Concave Up

Determine the exact value of sin⁻¹(3/7) - cos⁻¹(-3/7)

-π/2

Find the absolute maximum and absolute minimum values of the function f(t) = 2cos(t) + sin(2t) on [0, 2π].

Absolute Max: (3√3)/2 at x = pi/6 Absolute Min: -(3√3)/2 at x = (5pi)/6

Find the absolute maximum and absolute minimum values of the function f(x)=x↑(2/3) on [−1,8].

Absolute Max: 4 at x = 8 Absolute Min: 0 at x = 0

Let f(x) be the function given by f(x) = 4sin↑(3)(x) + 21sin↑(2)(x) − 24sin(x). Find its absolute maximum value and absolute minimum value on the interval [−π2, π2].

Absolute Max: 41 at t = -1 (x = -(π/2) Absolute Min: -(25/4) at x = 1/2 (x = π/6)

Find the absolute maximum and absolute minimum values of the function f(x)=(x↑(2)−1)↑(3) on [−1,3].

Absolute Max: 512 at x = 3 Absolute Min: -1 at x = 0

Find the absolute maximum and absolute minimum values of the function f(x)=xe↑(−x) on [−4,6].

Absolute Max: e↑(-1) at x = 1 Absolute Min: -4e↑(4) at x = -4

Find the absolute maximum and absolute minimum values of the function f(x) = ln(x↑(2) + x + 1) on [−1, 1].

Absolute Max: ln(3) at x = 1 Absolute Min: ln(3/4) at x = -(1/2)

What is the largest area of a right triangle whose hypotenuse has length 5?

Largest Area = 25/4

What is the largest area of the rectangle inscribed in an ellipse (x²/9) + (y²/4) = 1.

Largest area = 12

We have a function whose first derivative is given by the formula f′(x) = (x−1)↑(2)(x+3)↑(3). Find the local extrema and the inflection points of the function.

Local Extrema: local min at x = -3 inflection point(s): x = -(3/5), 1