Solutions to Homework 8, Mathematics 1

Problem 1 [6 points]: Do the detailed graphing (definition domain, intersections with theaxes, limits at ±∞, monotonicity, local and global extrema, convexity/concavity, inflection

points, sketch of graph) for the function f(x) = arctan(√

3x2

).

Solution: We begin by looking at the maximal domain of the function f . There is only a issuewith x = 0, where the argument of arctan becomes undefined. So we have Domain(f) = R\{0}.

Then it is nice to look at the limits at the “borders” of the intervals where f is defined. Inthis case, we should look at

limx→∞

f(x), limx→−∞

f(x), and limx→0±

f(x),

but perhaps it would be nicer to simply realize first that f is an even function, so we can justplot the graph for the positive numbers, and then just reflect it with respect to the y-axis. Wehave

limx→∞

f(x) = limx→∞

arctan

(√3

x2

)= arctan (0) = 0,

and

limx→0+

f(x) = limx→0−

f(x) = limx→0

arctan

(√3

x2

)=π

2.

Then even if the function above is not defined on x = 0, we can extend it to a continuousfunction by letting f(0) = π/2. But we will not worry about it here.

We see that the function is never 0, so the graph of f does not touch the x-axis. Since f isnot defined for x = 0, its graph also does not touch the y-axis. If we extend the function asabove, then the graph would have an intersection with the y-axis at (0, π/2) however.

So, let us study the monotonicity. To do that, since we are talking about a differentiablefunction, it is enough to study the derivative of f . Without further ado, we have

f ′(x) =

(arctan

(√3

x2

))′

=1

1 +(√

3x2

)2 · −2√

3

x3

= −2√

3 · x

x4 + 3

So f ′(x) is always negative for positive x and positive for negative x! This means that f ismonotone strictly increasing on (−∞, 0) and monotone strictly decreasing on (0,∞). It alsofollows that f has no local or global extrema. Note that if we extended the function by π

2in

0, then f would have a maximum (both local and global) in 0 (as f is increasing upto 0 andthen decreasing after 0). Minimum of f is not achieved, infimum of f equals 0 (as f → 0 forx→ ±∞).

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Figure 1. Graph of f from Problem 1.

Let us study the concavity of the function now, using the second derivative. We have

f ′′(x) =

(−2√

3 · x

x4 + 3

)′= −2

√3 · x

4 + 3− x · 4x3

(x4 + 3)2

= 6√

3 · x4 − 1

(x4 + 3)2

= 6√

3 · (x− 1)(x+ 1)(x2 + 1)

(x4 + 3)2

Therefore we know that f ′′ is positive on (−∞,−1) ∪ (1,∞), negative on (−1, 1) \ {0}, and 0on {−1, 1}. We come to this conclusion after studying the changes of sign as x varies in thefunctions (x − 1) and (x + 1). We then have that f in convex on (−∞, 1) and (1,∞) andconcave on (−1, 0) and (0, 1) (after the extension, even on (−1, 1)). The inflection points of fare at x = −1 and x = 1. All this information allows us to plot the function f (see Figure 1).

Problem 2: Let a (one-dimensional) displacement of a particle be given by the formula s(t) =√3t+ 4, t ≥ 0. Find time t0 ∈ (0, 4) such that the instantenous velocity at time t0 is equal

to the average velocity over the interval [0, 4]. Identify what this t0 is corresponding to in theLagrange’s Mean Value Theorem.

Solution: (By Dr. Katarina Bellova) The average velocity on the interval [0, 4] is equal to

vavg =s(4)− s(0)

4=

√3 · 4 + 4−

√3 · 0 + 4

4=

4− 2

4=

1

2.

The instantaneous velocity is equal to

v(t) = s′(t) =((3t+ 4)1/2

)′=

1

2(3t+ 4)−1/2 · 3 =

3

2√

3t+ 4.

3

Figure 2. A sketch of the graph of f from Problem 3. Only x ∈ [0, 2] should be considered.

If this is equal to vavg, we must have:

1

2=

3

2√

3t+ 4,

√3t+ 4 = 3,

3t+ 4 = 9,

t = 5/3.

On the other hand, indeed

v(5/3) =3

2√

3 · 5/3 + 4=

3

2√

9=

1

2= vavg

(alternatively, one can check that the manipulations before were equivalent, namely that we canalso get from the bottom to top: in particular, check that in the step where one takes squareroot, both sides are positive and later we don’t divide by 0). Hence, the t we were looking for is

t0 = 5/3. Since for this t0 we have s′(t0) = s(4)−s(0)4

, this t0 corresponds to the ξ in Lagrange’sMean Value Theorem applied on interval [0, 4] to the function s (as s satisfies the assumptionsof the theorem, we are guaranteed the existence of the ξ = t0).

Problem 3: A 2 meter piece of wire is cut into two pieces and one piece is bent into a squareand the other is bent into an equilateral triangle. Where should the wire be cut so that thetotal area enclosed by both is minimum and maximum?

Solution: We cut this wire into a piece of size x and a piece of size 2 − x. With the pieceof size x we make an equilateral triangle with side length x/3. And with the piece of size 2− xwe make a square with side length (2− x)/4. Using the Pythagorean theorem, or recalling thesine of π/3, we can calculate the triangle’s area to be

√3x2

36.

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Figure 3. Sketch of the layout of the cable in Problem 4.

The square’s area is simplyx2 − 4x+ 4

16.

Our job is now to study the maximum and the minimum of the function

f(x) =

√3x2

36+x2 − 4x+ 4

16

on the interval [0, 2]. We can do this if we remember how to study the quadratic function graphfrom school. Or we can differentiate and get

f ′(x) =

√3x

18+x− 2

8=

4√

3x+ 9x− 18

72.

And we have f ′(x) = 0 if and only if

x =18

4√

3 + 9≈ 1.13 ∈ [0, 2].

Furthermore, since f ′′ is constant and positive, we have that the above point is a point of aglobal minimum. The global maximum can then be found at the extremities, but we noticethat

f(0) =1

4, and f(2) =

√3

9=

1

3√

3<

1

4,

so the global maximum is attained at x = 0 (when we form only a square).

5

Problem 4 [4 points] A power house, P , is on one bank of a straight river 30m wide, and afactory, F , is on the opposite bank 40m downstream from P . The cable has to be taken acrossthe river, under water at a cost of 20 EUR/m. On land the cost is 10 EUR/m. What pathshould be chosen so that the cost is minimized?

Solution: Again an optimization problem. We should assume that the path taken by thecable consists of a union of straight line segments. It is easy to believe this, if you have somecurved (smooth) path between two points A and B, its length should be bigger than the lengthof the straight line segment between A and B. Without loss of of generality, assume that thepath looks like in Figure 3: first the cable crosses the river (under some angle) and then itcontinues on land. Other options (first going along the river and then crossing into F , or firstgoing along the river, then crossing along some angle and lastly going along the river on theother side) will give us the same costs, as long as the angle of crossing the river will be thesame.

Consider R, the point directly opposed to P on the other side of the river (I think the Germanword is gegenuber), and S, the point where the cable emerges from water on the side of thefactory. That is, the path of the cable will consist of the segments PS and SF , and we have

length(PR) = 30, length(RS) = x, length(SF ) = 40− x, length(PS) =√

302 + x2.

Our task is to minimize the function

20 · length(PS) + 10 · length(SF ) = 400− 10x+ 20√

302 + x2 = f(x),

for x ∈ [0, 40]. We have

f ′(x) = −10 + 202x

2√

900 + x2.

Therefore, f ′(x) = 0 if and only if

0 = −10 + 202x

2√

900 + x2

1 =2x√

900 + x2

900 + x2 = 4x2

x2 = 300

x = 10√

3.

This x indeed lies in [0, 40]. We also notice that

f ′′(x) =20√

900 + x2 − 20x 2x2√900+x2

900 + x2,

and

f ′′(10√

3) > 0.

Therefore, 10√

3 is a minimum (a global one since f ′ is only 0 at this point) and it determines(one of) the optimal path(s).

6

Problem 5 [4 × 3 points]: Compute the limits, possibly using l’Hospital rule (make sure tocheck it’s assumptions):

(a) limx→a

ax − xa

x− afor a > 0,

(b) limx→1

x1

1−x ,

(c) limx→0

(cosx)1

sin x ,

(d) limx→0

(1

x− cotg x

),

(e) limx→π

√1− tanx−

√1 + tan x

sin(2x),

(f) limx→e

ln(lnx)

sin(x− e),

(g) limx→0+

(1− x lnx)1√x

Hint: Some material from Wednesday’s lecture might be useful.

Solution:

(a): (By Dr Katarina Bellova)

limx→a

ax − xa

x− a“ 00”

= limx→a

(ax − xa)′

(x− a)′= lim

x→a

ax · ln a− axa−1

1

= aa · ln a− aaa−1 = aa(ln a− 1).

Notice we used a > 0 (and hence x > 0 near a) several times. Since the limit on the rightexists, l’Hospital rule is justified.

(b): (By Dr Katarina Bellova)

limx→1

x1

1−x = limx→1

eln x1−x = elimx→1

ln x1−x ,

limx→1

lnx

1− x“ 00”

= limx→1

(lnx)′

(1− x)′= lim

x→1

1/x

−1= −1, so

limx→1

x1

1−x = e−1 =1

e.

Since the limit on the right in the middle equality exists, l’Hospital rule is justified. Then weused the continuity of y 7→ ey.

(c): We have

limx→0

(cosx)1

sin x = limx→0

eln(cos x)

sin x

= exp

(limx→0

ln(cosx)

sinx

)“ 00”

= exp

(limx→0

− sinxcosx

cosx

)= e0

= 1.

7

(d):

limx→0

(1

x− cotg x

)= lim

x→0

(1

x− cosx

sinx

)= lim

x→0

(sinx− x cosx

x sinx

)“ 00”

= limx→0

(cosx− cosx+ x sinx

sinx+ x cosx

)“ 00”

= limx→0

(sinx+ x cosx

cosx+ cosx− x sinx

)= 0.

(e):

limx→π

√1− tanx−

√1 + tan x

sin(2x)

“ 00”

= limx→π

12√1−tanx

−1cos2 x

− 12√1+tanx

1cos2 x

2 cos(2x)

= −1

2.

(f): (By Dr Katarina Bellova)

limx→e

ln(lnx)

sin(x− e)“ 00”

= limx→e

1lnx

1x

cos(x− e)=

111e

cos 0=

1

e.

(g): First, notice that

(1) limx→0+

x lnx = limx→0+

lnx1x

“∞∞”= lim

x→0+

1x

− 1x2

= 0,

so the bracket is positive for x→ 0+. We thus have

limx→0+

(1− x lnx)1√x = lim

x→0+exp

(ln(1− x lnx)√

x

)= exp

(limx→0+

ln(1− x lnx)√x

).

Because of (1), we see that the above limit is also of the form 00. Therefore we may use l’Hospital

rule to prove that the above limit is equal to

exp

(limx→0+

− lnx−11−x lnx

12√x

)= exp

(limx→0+

2−√x lnx−

√x

1− x lnx

)= e0

= 1.

Here we used that limx→0+

√x lnx = 0 - this result can be obtained similarly as (1), or one can

just refer to Ex. 2.2.20.

Problem 6 [3 bonus points]: Consider the function

f(x) =

{e−

1x2 if x 6= 0,

0 if x = 0.

Compute f (n)(0) for all n ∈ N.

8

Solution: (By Dr. Katarina Bellova) First, notice that

limx→0

e−1x2

y= 1x2= lim

y→+∞e−y = lim

y→+∞

1

ey= 0,

so f is continuous. Next,

f ′+(0) = limx→0+

e−1x2

x

y= 1x= limy→+∞

y

ey2“∞∞”= lim

y→+∞

1

2yey2= 0,

f ′−(0) = limx→0−

e−1x2

x

y= 1x= limy→−∞

y

ey2“∞∞”= lim

y→−∞

1

2yey2= 0,

(we used l’Hospital rule...) so f ′(0) = 0.To compute f ′′(0), we also need to know f ′(x) for x 6= 0:

f ′(x) =2

x3e−

1x2 .

Then

f ′′(0) = limx→0+

2x3e−

1x2

x

y= 1x2= lim

y→+∞

2y2

ey“∞∞”= lim

y→+∞

4y

ey“∞∞”= lim

y→+∞

4

ey= 0.

To compute f (n)(0) for general n, we first express f (n)(x) for x 6= 0. Trying out a few morederivatives suggest that for n ≥ 0 and x 6= 0, f (n)(x) has the form

(2) f (n)(x) = Pn

(1

x

)e−

1x2 ,

where Pn(y) = an3ny3n + an3n−1y

3n−1 + . . . + an1y + an0 is a polynomial of order 3n (actually, thelowest nonzero term will be ann+2y

n+2, but we will not need this). We prove statement (2) byinduction. For n = 0, this is true with P0(y) = 1 being of order 0. (We even checked thestatement for n = 1 with P1(y) = 2y3.) Now assume (2) holds for n = k (k ∈ N0), we willprove it for n = k + 1. Differentiating (2) for n = k gives

f (k+1)(x) = (f (k))′(x) =

(Pk

(1

x

)e−

1x2

)′=

(Pk

(1

x

))′e−

1x2 + Pk

(1

x

)(e−

1x2

)′=

(ak3k

1

x3k+ ak3k−1

1

x3k−1+ . . .+ ak1

1

x+ ak0

)′e−

1x2

+

(ak3k

1

x3k+ ak3k−1

1

x3k−1+ . . .+ ak1

1

x+ ak0

)2

x3e−

1x2

=

(−3kak3k

1

x3k+1− (3k − 1)ak3k−1

1

x3k− . . .− ak1

1

x2

)e−

1x2

+

(2ak3k

1

x3k+3+ 2ak3k−1

1

x3k+2+ . . .+ 2ak1

1

x4+ 2ak0

1

x3

)e−

1x2

= Pk+1

(1

x

)e−

1x2 ,

where

Pk+1(y) =2ak3ky3k+3 + 2ak3k−1y

3k+2 + (2ak3k−2 − 3kak3k)y3k+1 + . . .

+ (2ak0 − 2ak2)y3 − ak1y2,

which is a polynomial of order 3(k + 1), so we proved (2) for n = k + 1 and hence for anyn ∈ N0.

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Next, one shows that for any n ∈ N0,

(3) limx→0

1

xne−

1x2 = 0.

For n even, this can be done in the same way as when we computed (3) for n = 0 and n = 4above (when checking continuity of f at 0 and computing f ′′(0)): after the substitution y = 1

x2,

we can use l’Hospital’s rule n/2-times. For n odd, we need to split the limit into limx→0− andlimx→0+ , and proceed as when we computed (3) for n = 1 (when computing f ′(0)).

After (2) and (3) are proved, f (n)(0) = 0 follows easily.