18.01 Solutions, Problem Set 3

• §2.6: #2,4,7

Verify that the function f(x) satisfies the hypotheses of the Mean Value Theorem on the given interval, and find all points c whose existence is guaranteed by the theorem.

2. f(x) = sin x, [0, 2π].

Solution: Since f(x) is continuous and differentiable (everywhere), it satisfies the hypotheses of the MVT. Since sin(2π) = sin(0) = 0 and f′(x) = cos x, we are looking for points c satisfying

cos c = \frac{sin(2π) − sin 0}{2π} = 0.

This happens for c = π/2 and c = 3π/2.

4. f(x) = x^{3} + 1, [1, 2].

Solution: Again f(x) is continuous and differentiable. We have f′(x) = 3x^{2}, f(2) = 9, f(1) = 2. Solving 3c^{2} = 7 gives c = ±\sqrt{7/3}, but only c = \sqrt{7/3} is in the given interval.

7. f(x) = \sqrt{x + 1}, [0, 3].

Solution: The function is defined and continuous for x ≥ −1, and its derivative f′(x) = \frac{1}{2}(x + 1)^{−1/2} is defined for x > 1. So f(x) is continuous and differentiable on [0, 3]. Setting

\frac{1}{2}(c + 1)^{−1/2} = \frac{f(3) − f(0)}{3} = \frac{1}{3}

gives (c + 1)^{−1/2} = 2/3, so c + 1 = (2/3)^{−2} = 9/4, and the solution is c = 5/4.

• §2.6: #10

If f(x) is an arbitrary quadratic polynomial, that is, if f(x) = Ax^{2} + Bx + C(A ≠ 0), show that the point c whose existence is guaranteed by hte MVT is the midpoint of the interval [a, b].

Solution: Since f′(x) = 2Ax + B, the point c satisfies

2Ac + B = \frac{(Ab^{2}+Bb+C)−(Aa^{2}+Ba+C)}{b−a}
= \frac{A(b^{2}−a^{2})+B(b−a)}{b−a}
= \frac{A(b+a)(b−a)+B(b−a)}{b−a}
= A(b + a) + B.

So c = (b + a)/2, which is the midpoint of the interval.

• §4.1: #2,8,13

Sketch the graphs of the following functions by using the first derivative and the methods of this section; in particular, find the intervals on which each function is increasing and those on which it is decreasing, and locate any maximum or minimum values it may have.

2. y = 2 + x − x^{2}.

Solution: The first derivative of f(x) = 2+x−x^{2} is f′(x) = 1−2x. The critical points are the solutions of 1 − 2x = 0, so x = \frac{1}{2}. We consider whether f(x) is increasing or decreasing on the intervals:

(−∞, 1/2) and (1/2, ∞).

For x < \frac{1}{2}, f(x) is increasing since f′(x) > 0. For x > \frac{1}{2}, f(x) is decreasing since f′(x) < 0. Therefore there is a maximum at x = 1/2; the maximum value is f(1/2) = 9/4. The zeros occur when 2 + x − x^{2} = (1 + x)(2 − x) = 0, so x = −1, 2.

8. y = x^{4} − 2x^{2} + 1.

Solution: The derivative of f(x) = x^{4}−2x^{2}+1 is f′(x) = 4x^{3}−4x. The critical points are the solutions of

4x^{3} − 4x = 4x(x + 1)(x − 1) = 0,

so x = 0, ±1. We consider whether f(x) is increasing or decreasing on the intervals with those endpoints (see the table). For x < −1, f′(x) < 0 (the factors x, x − 1 and x + 1 are all negative), so f(x) is decreasing. For −1 < x < 0, f′(x) > 0 (the factor x + 1 is positive, the other two are negative), so f(x) is increasing. Similarly f(x) is decreasing on (0, 1) and increasing on (1, ∞).

\begin{matrix}(−∞, −1) & (−1, 0) & (0, 1) & (1, ∞) \\ ↘ & ↗ & ↘ & ↗ \end{matrix}

We therefore have minima at x = ±1 (where the value is 0), and a maximum at x = 0 (where the value is 1). Since

x^{4} − 2x^{2} + 1 = (x^{2} − 1)^{2} = (x + 1)^{2}(x − 1)^{2},

the only zeros are at x = ±1.

13. y = \frac{1}{x^{2} + x}.

Solution: The function f(x) = (x^{2} +x)^{−1} is not defined if x^{2} +x = 0, which happens when x = −1, 0. The derivative is

f′(x) = −(x^{2} + x)^{−2}(2x + 1) = \frac{−(2x + 1)}{(x^{2} + x)^{2}},

which is zero when 2x + 1 = 0, so when x = −1/2. We consider f(x) on the intervals with endpoints −1, −1/2, 0. Noting that the denominator in the above expression for f′(x) is positive, we get:

\begin{matrix}(−∞, −1) & (−1, −1/2) & (−1/2, 0) & (0, ∞) \\ ↗ & ↗ & ↘ & ↘ \end{matrix}

Therefore there is a maximum at x = −1/2, where the value is −4. Note that f(x) is never 0. Note also that f(x) → 0 as x → −∞ or as x → ∞. Finally the denominator → 0 in the expression for f(x) as x → −1 or 0, and the behavior near those points can then be determined from the table above.

• §4.1: #17

The function f(x) = x^{3} +x−1 being a third-degree polynomial, obviously crosses the x-axis (why?) and therefore has at least one zero. By examining f′(x), show that this function has only one zero. Show similarly that f(x) = 7x^{131} + 11x^{73} + x − 500 has one and only one zero.

Solution: Since f(x) = x^{3} + x − 1 has degree 3 (and positive leading coefficient), f(x) → ∞ as x → ∞ < and f(x) → −∞ as x → −∞. So the graph must cross the axis. Since f′(x) = 3x^{2} + 1 is always positive, the function is always increasing, so the graph crosses the axis only once.

Similarly f(x) = 7x^{131} + 11x^{73} + x − 500 has odd degree, so at least one zero. Since f′(x) = 917x^{130} +803x^{72} +1 is positive, there is only one zero.

• §4.1 #23

Find the critical points and corresponding critical values for the function y = cos(2x) + 2 cos x on the interval [0, 2π]. Sketch the graph.

Solution: We have

y′ = −2 sin(2x) − 2 sin x = −2(2 sin x cos x + sin x) = −2 sin x(2 cos x + 1),

which is zero when sin x = 0 (so x = 0, π, 2π), or cos x = −1/2 (so x = 2π/3, 4π/3).

\begin{matrix}(0, 2π/3) & (2π/3, π) & (π, 4π/3) & (4π/3, 2π) \\ ↘ & ↗ & ↘ & ↗ \end{matrix}

The critical values are given by

x = 0 2π/3 π 4π/3 2π
f(x) = 3 −3/2 −1 −3/2 3
max  min max min max.

• §4.1 #30

Find all critical points for each fo the following functions, use the first derivative to decide whether each of the corresponding critical values is a maximum, a minimum, or neither, and sketch the graphs.

a) y = x + sin x;

Solution: Then y′ = 1+cos x. This is 0 when cos x = −1, so when x = π plus an integer multiple of 2π. That is, x = π + 2πn for all integers n. The critical value is π +2πn at the critical point π +2πn. But since y′ > 0 otherwise, there are no maxima or minima; y is always increasing. b) y = sin^{2} 2x;

Solution: Then y′ = 4 sin(2x) cos(2x) = 2 sin(4x). This is 0 when when x is a multiple of π/4. Since y′ > 0 when x is in [0, π/4], [π/2, 3π/4], etc., and y′ < 0 for x in [−π/4, 0], [π/4, π/2], etc., the critical values alternate between minima (critcal value 0) at multiples of π/2 and maxima (critcal value 1) at the rest.

• §4.2: 5,8,11

For each of the following, locate the points of inflection, find the intervals on which the curve is concave up and those on which it is concave down, and sketch.

5. y = x^{4} + 2x^{3} + 1.

Solution: Taking the derivative, we get y′ = 4x^{3} + 6x^{2} and the critical points are x = −3/2, 0.

\begin{matrix}(−∞, −3/2) & (−3/2, 0) & (0, ∞) \\ ↘ & ↗ & ↗ \end{matrix}

Taking the second derivative, we get y′′ = 12x^{2} + 12x. The possible inflection points are x = −1, 0. Since y′′ > 0 for x < −1 and x > 0, the graph is concave up on those intervals, and concave down on (−1, 0). The points of inflection are x = −1,y= 0, andx=0, y = 1.

\begin{matrix}(−∞, −1) & (−1, 0) & (0, ∞) \\ ∪ & ∩ & ∪ \end{matrix}

8. y = 3x^{5} − 5x^{4}.

Solution: We find y′ = 15x^{4} − 20x^{3} and the critical points are x = 0, 4/3.

\begin{matrix}(−∞, 0) & (0, 4/3) & (4/3, ∞) \\ ↗ & ↘ & ↗ \end{matrix}

The second derivative is y′′ = 60x^{3} − 60x^{2}. The possible inflection points are x = 0, 1. The concavity is given by:

\begin{matrix}(−∞, 0) & (0, 1) & (1, ∞) \\ ∩ & ∩ & ∪ \end{matrix}

So the only point of inflection is x = 1, y = −2.

11. y = \frac{4x^{2}}{x^{2} + 3}.

Solution: Then y′ = \frac{4(2x(x^{2} + 3) − 2x(x^{2}))}{(x^{2} + 3)^{2}} = \frac{24x}{(x^{2} + 3)^{2}} so the

only critical point is x = 0. Note that the denominator is never zero, so y is always defined.

\begin{matrix}(−∞, 0) & (0, ∞) \\ ↘ & ↗ \end{matrix}

The second derivative is

\begin{aligned}y′′ & = 24(x^{2} + 3)^{−2} − 96x^{2}(x^{2} + 3)^{−3} \\ & = \frac{24(x^{2} + 3) − 96x^{2}}{(x^{2} + 3)^{3}} = \frac{72(1 − x^{2})}{(x^{2} + 3)^{3}}. \end{aligned}

The possible inflection points are x = ±1. The concavity is given by:

\begin{matrix}(−∞, −1) & (−1, 1) & (1, ∞) \\ ∩ & ∪ & ∩ \end{matrix}

So the inflection points are x = ±1, y = 1.

• §4.2 #15

Sketch the graph of a function f(x) defined for all x such that a) f(x) > 0, f′(x) > 0, and f′′(x) > 0.

Solution: f(x) = e^{x} works.

b) f′(x) < 0 and f′′(x) < 0.

Solution: f(x) = −e^{x} works.

• §4.2 #19

Find the value of a that makes y = x^{3} − ax^{2} + 1 have a point of inflection at x = 1.

Solution: Taking the first and second derivative, we get y′ = 3x^{2} − 2ax and y′′ = 6x − 2a. To have 6x − 2a = 0 at x = 1, we need 6 − 2a = 0, so a = 3. Then y′′ = 6x − 6 changes sign at x = 1, so there is a point of inflection.

• §4.2 #26

Show that the graph of y = sin x is concave down when it is above the x-axis and concave down when it is below the x-axis.

Solution: We have y′ = cos x and y′′ = − sin x. So y′′ is negative when y is positive, and y′′ is positive when y is negative. That is, y′′ = −y.

• §4.3: #4

Show that the rectangle with minimum perimeter for a given area is a square.

Solution: Let x be the length of the base and y the height of the rectangle. So we want to minimize the perimeter P = 2x+2y subject to the constraint that A = xy is constant. Solving the latter equation for y gives y = Ax^{−1}, so we want to minimize

P = f(x) = 2x + 2Ax^{−1}.

Setting the derivative f′(x) = 2 − 2Ax^{−2} equal to 0 gives 2 − 2Ax^{−2} = 0. Solving 2 = 2Ax^{−2} by multiplying through by x^{2}/2 gives x^{2} = A, so x = \sqrt{A} (since x > 0). This yields a minimum for P since f′′(x) = 4Ax^{−3} > 0. The rectangle with x = \sqrt{A} is a square since then y = Ax^{−1} = \sqrt{A} = x.

• §4.3 #8

A certain poster requires 96 in^{2} for the printed message and must have 3-in margins at the top and bottom and a 2-in margin on each side. Find the overall dimensions of the poster if the amount of paper used is a minimum.

Solution: Let x be the width (in inches) of the printed area and let y be the height (in inches). The dimensions of the poster are then x + 4 and y + 6. We wish to minimize the amount of paper: A = (x + 4)(y + 6), subject to the constraint xy = 96. Solving for y and substituting into the formula for A gives

A = f(x) = (x + 4)(96x^{−1} + 6) = 6(x + 20 + 64x^{−1}).

So f′(x) = 6(1 − 64x^{−2}) which is 0 when x^{2} = 64, so x = 8 (since x > 0). This gives a minimum since f′′(x) = 768x^{−3} > 0. We have y = 12, so the the poster should be 8 + 4 = 12 inches wide and 12 + 6 = 18 inches high.

• §4.3 #10

A new branch bank is to have a floor area of 3500 ft^{2}. It is to be a rectangle with three solid brick walls and a decorative glass front. The glass costs 1.8 times as much as the brick wall per linear foot. What dimensions of the building will minimize the cost of materials for the walls and front?

Solution: Let x be the width (in feet) of the front and y the width (in feet) of the sides. The total cost is then C = 2.8x + 2y, and the area is xy = 3500. So

C = f(x) = 2.8x + 7000x^{−1}.

Then f′(x) = 2.8 − 7000x^{−2}, which is 0 when x^{2} = 2500, so x = 50 and y = 70. This gives a minimum since f′′(x) = 14000x^{−3} > 0. The dimensions of the building are 50 ft (front) by 70 ft (sides).

• §4.3 #15

Find the area of the largest rectangle with lower base on the x-axis and upper vertices on the parabola y = 27 − x^{2}.

Solution: If the vertex in the first quadrant is at the point (x, y), then the area of the rectangle is A = 2xy. Substituting y = 27 − x^{2} gives A = f(x) = 54x − 2x^{3}, so f′(x) = 54 − 6x^{2}. This is 0 when x = 3 (since x > 0). This gives a maximum since f′′(x) = −12x^{3} < 0 if x > 0. The value of y is 18, and the area is 108.

• §4.3 #28

A fence a feet high is b feet from a high burning building. Find the length of the shortest ladder that will reach from the ground across the top of the fence to the building.

Solution: Let θ be the angle of inclination of the ladder. Then the dis-

tance from the base of the ladder to the building is b + a\frac{cos θ}{sin θ}, and the length of the ladder is

L = f(θ) = (b + a\frac{cos θ}{sin θ})\frac{1}{cos θ} = \frac{b}{cos θ} + \frac{a}{sin θ} .

So

f′(θ) = \frac{b sin θ}{cos^{2} θ} + −\frac{a cos θ}{sin^{2} θ} = \frac{b sin^{3} θ − a cos^{3} θ}{cos^{2} θ sin^{2} θ} .

So f′(θ) = 0 when tan^{3} θ = a/b. This gives a minimum with θ between 0 and π/2, since then

f′′(θ) = \frac{b(1 + sin^{2} θ)}{cos^{3} θ} + \frac{a(1 + cos^{2} θ)}{sin^{3} θ}

is positive. Therefore

tan θ = \frac{a^{1/3}}{b^{1/3}} , sin θ = \frac{a^{1/3}}{(a^{2/3} + b^{2/3})^{1/2}} , cos θ = \frac{b^{1/3}}{(a^{2/3} + b^{2/3})^{1/2}} .

The length of the ladder is therefore

L = a^{2/3}(a^{2/3} + b^{2/3})^{1/2} + b^{2/3}(a^{2/3} + b^{2/3})^{1/2} = (a^{2/3} + b^{2/3})^{3/2}.

• §4.4 #4

Find the height of the cone of maximum volume that can be inscribed in a sphere of radius R.

Solution: Let h be the height of the cone base and r the radius of its base. The volume of the cone is \frac{π}{3} r^{2}h. To relate r and h, note the right triangle

with sides of length r, h − R and hypotenuse R. So R^{2} = r^{2} + (h − R)^{2}, which gives r^{2} = 2Rh − h^{2}. Substituting this for r^{2} in the formula for the volume gives

V = f(h) = \frac{π}{3} (2Rh^{2} − h^{3}).

Then f′(h) = \frac{π}{3} (4Rh − 3h^{2}) = 0 when h = 4R/3 (ruling out h = 0). This

gives a maximum since f′′(h) = \frac{π}{3} (4R−6h) which is < 0 when h = 4R/3.

• §4.4 #7

A cylindrical can without a top is to be made from a specified weight of sheet metal. Find the ratio of the height to the diameter of the base when the volume of the can is to be the greatest.

Solution: Let h be the height and r the radius of the base. The volume is V = πr^{2}h. r and h are related because the weight of the metal is fixed, and therefore so is the total area A = πr^{2} + 2πrh. Solving for h gives h = \frac{1}{2}(\frac{A}{πr} − r). Substituting gives

V = f(r) = \frac{π}{2} (Ar/π − r^{3}).

Then f′(r) = \frac{π}{2} (A/π − 3r^{2}), which is 0 when r = \sqrt{A/3π} (assuming r > 0). This gives a maximum since f′′(r) = −3πr is negative if r > 0. The height is then

h = \frac{1}{2}(3\sqrt{A/3π} − \sqrt{A/3π}) = \sqrt{A/3π} = r,

so the height is 1/2 the diameter of the base.

• §4.4 #9

Draw a reasonably good sketch of y = \sqrt{x} and mark the point on the graph that seems to be closest to the point (\frac{3}{2}, 0). Then calculate the coordinates of this closest point. Hint: Minimize the square of the distance from the point (\frac{3}{2}, 0) to the point (x, \sqrt{x}).

Solution: The square of the distance between the points is

f(x) = (x − \frac{3}{2})^{2} + (\sqrt{x})^{2} = x^{2} − 2x + \frac{9}{4}.

So f′(x) = 2x − 2, which is 0 when x = 1. This gives a minimum since f′′(x) = 2 > 0. The coordinates of the closest point are (1, 1).

• §4.4 #22

There is a refinery at a point A on a straight highway and an oil well at a point B which can be reached by traveling 5 mi along the highway to a point C and then 12 mi across country perpendicular to the highway. If a pipeline is built from A to B, it costs k times as much to build it across country as along the highway, because of the difficult terrain. The line will be built either directly from Ato B or along the highway to a point P part of the way toward C and then across country to B, whichever is cheaper. Decide on the cheapest route

a) if k = 3; Solution: Snell’s law applies to give sin α = 1/k where α is the angle between the pipeline and the line perpendicular to the road. Since 1/3 < 5/13, the cost is minimized by following the road until the angle satisfies sin α = 1/3.

b) if k = 1/2;

Solution: Since 1/2 > 5/13, the cost is minimized by a direct pipeline.

c) What is the largest value of k for which it is cheapest to build the pipeline directly from A to B?

Solution: This occurs when 1/k = 5/13, so k = 13/5 = 2.6.

• §4.4 #31

A spider at a corner S of the ceiling of a cubic room 8 ft on each side wishes to catch a bug at the opposite corner B on the floor. The spider, who must walk on the ceiling, the walls, or the floor, wishes to find the shortest path to the bug. Find a shortest path

a) by using calculus;

Solution: With x as in the picture, the distance is

d = f(x) = \sqrt{8^{2} + (8 − x)^{2}}+\sqrt{8^{2} + x^{2}} = \sqrt{x^{2} − 16x + 128}+\sqrt{x^{2} + 64}.

So f′(x) = \frac{2x − 16}{2\sqrt{x^{2} − 16x + 128}} + \frac{2x}{2\sqrt{x^{2} + 64}}, which is zero when

(x^{2} − 16x + 64)(x^{2} + 64) = x^{2}(x^{2} − 16x + 128).

This reduces to −1024x + 4096 = 0, so x = 4. This gives a minimum since f′′(x) = 128[(x^{2} − 16x + 128)^{−3/2} + (x^{2} + 64)^{−3/2}], which is negative.

b) without calculus, by merely thinking.

Solution: Flattening the surfaces onto a plane, we see that a straight path from S to B bisects the edge between the ceiling and wall.