M ATH 150: S EMINAR #11 Q UESTIONS
S OLUTIONS
Content:
Section 4.5: Summary of Curve Sketching
Section 4.7: Optimization
Section 4.8: Newton’s Method
Questions:
1. Sketch the graph of
f (x) = x2 e−x .
On your graph clearly indicate and label all intercepts, asymptotes, local extrema, and inflection points.
Solution.
• Since f is a product of a polynomial and an exponential function its domain is the set of all real
numbers.
• We note that the x-intercept (and the y-intercept) is the point (0, 0).
• From, for all x ∈ R,
f (−x) = (−x)2 e−(−x) = x2 ex
we conclude that f is neither odd nor even function.
• We use the fact that
lim ex = lim e−x = ∞
x→∞
x→−∞
to conclude that
x2
=0
x→∞ ex
lim f (x) = lim x2 e−x = lim
x→∞
x→∞
and
lim f (x) = lim x2 e−x = ∞
x→−∞
x→−∞
to conclude that the line y = 0 is a horizontal asymptote (when x → ∞.)
• Next we find the first derivative:
f ′ (x) = 2xe−x − x2 e−x = xe−x (2 − x).
Since e−x > 0 for all x ∈ R, it follows that the sign of f ′ (x) depends only on the product x(2 − x).
Hence
0
2
x
− 0 + + +
2−x + + + 0 −
f ′ (x) − 0 + 0 −
f (x) ↘
↗
↘
and we conclude that the function is increasing on the interval (0, 2) and decreasing on the intervals
(−∞, 0) and (2, ∞). By the First Derivative Test, there is a local minimum at the point (0, f (0)) =
(0, 0) and a local maximum at the point (2, f (2)) = (2, 4e−2 ).
• Finally we find the second derivative:
f ′′ (x) = 2e−x − 2xe−x − 2xe−x + x2 e−x = e−x (2 − 4x + x2 ) .
Hence
f ′′ (x) = 0 ⇔ x2 − 4x + 2 = (x − 2)2 − 2 = (x − 2 −
Hence
√
x − 2 + √2
x−2− 2
f ′′ (x)
f (x)
√
2− 2
−
0
−
−
+
0
∪
+
−
−
∩
√
√
2+ 2
+
0
0
2)(x − 2 +
+
+
+
∪
√
2) = 0.
M ATH 150: S EMINAR #11 Q UESTIONS
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√
√
and we conclude that the function
is concave
down on the interval (2 − 2, 2 + 2) and
√
√ concave
√ up
√
on the intervals
(−∞,
√
√ 2 − 2) and (2 + 2, ∞). There are two inflection points(2 − 2, f (2 − 2))
and (2 + 2, f (2 + 2)).
• The graph of the function f (x) look like this:
Figure 1: f (x) = x2 e−x
2. Consider the function f (x) = x2/3
5
−x .
2
(a) Explain why f is continuous for x ∈ (−∞, ∞).
(b) Determine the behaviour of f as x → ±∞.
(c) Find f ′ (x), determine the intervals of increase and decrease of f . Identify the locations of the local
extrema of f and classify them as local maxima or minima.
(d) Find f ”(x), determine the concavity and identify the inflection points of f .
(e) Sketch the graph of the function f . On your graph clearly indicate and label all intercepts, local
extrema, and inflection points.
Solution.
(a) Explain why f is continuous for x ∈ (−∞, ∞).
Solution: Note that the function f is the product of a power function y = x2/3 and a linear function
5
y = − x. Both of these functions are defined for all real numbers. Hence they are continuous
2
there. Therefore, the function f , as their product, is continuous on (−∞, ∞).
(b) Determine the behaviour of f as x → ±∞.
Solution: From lim x
x→±∞
2/3
= ∞ and lim
x→±∞
5
−x
2
= ∓∞ we conclude that lim x
x→±∞
2/3
5
−x
2
=
∓∞.
(c) Find f ′ (x), determine the intervals of increase and decrease of f . Identify the locations of the local
extrema of f and classify them as local maxima or minima.
Solution: From
2
5
5 − 2x − 3x
5(1 − x)
√
√
f ′ (x) = x−1/3
− x − x2/3 =
=
3
3
2
3 x
33x
M ATH 150: S EMINAR #11 Q UESTIONS
S OLUTIONS
we observe that x = 0 and x = 1 are critical numbers of the function f . From the table
1−x
√
3
x
′
f (x)
+
−
−
f (x)
↘
0
+
0
not
defined
local
minimum
1
0
+
0
+
+
+
↗
−
+
−
local
↘
maximum
we conclude that the function f is decreasing on (−∞, 0) and (1, ∞) and increasing on (0, 1). There
is local minimum at the point (0, f (0)) = (0, 0) and local maximum at the point (1, 3/2).
(d) Find f ”(x), determine the concavity and identify the inflection points of f .
5
Solution: From f ′ (x) = · (x−1/3 − x2/3 ) it follows that
3
1 −4/3 2 −1/3
5 1 + 2x
5
.
− x
=− · √
f ”(x) = · − x
3
3
3
3
9
x4
From the table
1 + 2x
5
− √
3
9 x4
f ”(x)
−
−
+
f (x)
⌣
−1/2
0
−
0
+
−
−
inflection ⌢
point
0
+
+
0
−
not
−
defined
⌢
we conclude that the function is concave upward
on (−∞,
−1/2) and concave downward on (−1/2, 0)
1 3
and (0, ∞). There is an inflection point at − , √
.
2 34
(e) Sketch the graph of the function f . On your graph clearly indicate and label all intercepts, local
extrema, and inflection points.
Solution:
3. Sketch the graph of
f (x) = x3 e−x+5 .
On your graph clearly indicate and label all intercepts, local extrema, and inflection points.
Solution. The domain of the function f is the set of all real numbers. Also
f (x) = x3 e−x+5 = 0 ⇔ x = 0
so the origin is the only x− and y− intercept.
From
lim f (x) = lim
x→+∞
x3
x→+∞ ex−5
=0
and
lim f (x) = lim x3 e−x+5 = −∞
x→−∞
x→+∞
we conclude that the line y = 0 is the horizontal asymptote when x → +∞.
Since the first derivative is given by
f ′ (x) = 3x2 e−x+5 − x3 e−x+5 = x2 (3 − x)e−x+5
M ATH 150: S EMINAR #11 Q UESTIONS
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5
2 −x
Figure 2: f (x) = x2/3
we conclude that the critical numbers are x = 0 and x = 3. From the table
0
+ 0
+
+
+ 0
↗ 0
x2
3−x
e−x+5
f ′ (x)
f (x)
3
+
+ 0
+
+ 0
↗ 0
+
−
+
−
↘
we see that f is increasing on the interval (−∞, 3) and decreasing on the interval (3, ∞). By the first
derivative test, there is an absolute maximum at (3, f (3)) = (3, 27e2 ).
The second derivative is given by
f ”(x) = 2x(3 − x)e−x+5 − x2 e−x+5 − x2 (3 − x)e−x+5 = x(6 − 6x + x2 )e−x+5 .
Hence
f ”(x) = x(x − (3 −
√
3))(x − (3 +
From the table
x
√
x − (3 − √3)
x − (3 + 3)
f ”(x)
f (x)
0
− 0
−
−
− 0
∩
3−
+
−
−
+
∪
√
0
0
√
3))e−x+5 e−x+5 .
3
3+
+
+
−
−
∩
√
0
0
3
+
+
+
+
∪
M ATH 150: S EMINAR #11 Q UESTIONS
S OLUTIONS
√
we see that the graph of the function f is concave√downward on√the intervals (−∞, 0) and (3 − 3, 3 +
√
3) and
the intervals
√ concave
√ upward on √
√ (0, 3 − 3) and (3 + 3, ∞). The inflection points are (0, 0),
(3 − 3, f (3 − 3)), and (3 + 3, f (3 + 3)).
The graph of the function f is given below.
Figure 3: f (x) = x3 e−x+5
4. Sketch the graph of
y = 4x1/3 + x4/3 .
On your graph clearly indicate and label all intercepts, local extrema, and inflection points.
Solution. Let f (x) = 4x1/3 + x4/3 .
(a) The domain of f is the set of all real numbers.
(b) From f (0) = 0 it follows that the origin is both an x and the y intercept. From
f (x) = 4x1/3 + x4/3 = x1/3 (4 + x)
we get that that (−4, 0) is another x intercept.
(c) From
lim f (x) = ∞
x→±∞
we conclude that there is no horizontal asymptote. Since f is continuous on its domain, there is no
vertical asymptote. From
x1/3 (4 + x)
4
f (x)
1/3
= lim
= lim x
+ 1 = ±∞
lim
x→±∞ x
x→±∞
x→±∞
x
x
we conclude that there is no slant asymptote.
(d) From
f ′ (x) =
4 −2/3 4 1/3
4
4 1+x
·x
+ ·x
= · x−2/3 (1 + x) = · √
3
3
3
3 3 x2
we conclude that x = −1 and x = 0 are critical numbers.
M ATH 150: S EMINAR #11 Q UESTIONS
Since, for x ̸= 0,
√
3
S OLUTIONS
x2 > 0 we conclude that the sign of f ′ (x) depends only on the sign of 1 + x. Thus:
x
1+x
f ′ (x)
f (x)
−1
0
−
0
+
+
+
−
0
+
undefined
+
↘ local minimum ↗ inflection point ↗
Therefore f is increasing on (−1, 0) and (0, ∞) and decreasing on (−∞, −1). By the first derivative test, there is a local minimum at the point (−1, f (−1)) = (−1, −3) and an inflection point at
(0, f (0)) = (0, 0). Also we see that there is a vertical tangent line at (0.0).
(e) From
f ′ (x) =
4 −2/3
· x
+ x1/3
3
it follows that
f ”(x) =
4
2
1
4
4 x−2
.
· − · x−5/3 + · x−2/3 = · x−5/3 (−2 + x) = · √
3
3
3
9
9 x 3 x2
As before we conclude that the sign of f ”(x) depends only on the sign of x−2
x . Thus:
x
x−2 −
x
−
f ”(x) +
f (x)
∪
0
−
−
0
+
undefined
−
inflection point ∩
2
0
+
+
+
0
+
inflection point ∪
Therefore f is concave upward
√ on (−∞, 0) and (2, ∞) and concave downward on (0, 2). It follows
that (0, 0) and (2, f (2)) = (2, 6 3 2) are inflection points.
(f) Graph:
Figure 4: f (x) = 4x1/3 + x4/3
5. Find an equation of the slant asymptote. Do not sketch the curve.
M ATH 150: S EMINAR #11 Q UESTIONS
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x2 + 1
x+1
1
(b) y = 1 + x + e−x .
2
(a) y =
Solution.
(a) From
y=
x2 + 1
x2 − 1 + 2
x2 − 1
2
2
=
=
+
=x−1+
x+1
x+1
x+1
x+1
x+1
and the fact that
2
lim
x→∞ x + 1
=0
x2 + 1
is getting closer and closer to the line y = x−1
x+1
when x → ±∞. Hence, the line y = x − 1 is the slant asymptote to the graph of the given function.
we conclude that the graph of the function y =
(b) Observe that
lim e−x = 0
x→∞
and conclude that for very large positive values of x
1
y ≈ 1 + x.
2
1
Hence, the line y = 1 + x is the slant asymptote to the graph of the given function (when x → ∞.)
2
6. An open-topped cylindrical pot is to have volume 250 cm3 .The material for the bottom of the pot costs 4
cents per cm2 ; that for its curved side costs 2 cents per cm2 . What dimensions will minimize the total
cost of this pot?
Solution. Let h be the height of the pot and let r be the radius of its bottom. Then the volume of the
250
pot is given by V = r2 hπ = 250 cm3 which implies that h = 2 . The area of the bottom of the pot is
r π
B = r2 π cm2 and its cost is 4r2 π cents. The area of the curved side is
S = 2rπ · h = 2rπ ·
and its cost is 2 ·
250
500
=
cm2
r2 π
r
1000
500
=
cents. Hence the total cost of the pot is
r
r
C = C(r) = 4r2 π +
1000
,r > 0
r
dC
1000
8(r3 π − 125)
dC
= 8rπ − 2 =
it follows that
= 0 if r3 π − 125 = 0. Hence C = C(r)
dr
r
r2
dr
5
has only one critical number r = √
. Since
3
π
cents. From
dC
5
> 0 ⇔ r3 π − 125 > 0 ⇔ r > √
3
dr
π
and
dC
5
< 0 ⇔ r3 π − 125 < 0 ⇔ r < √
3
dr
π
5
it follows that the function C obtains a local minimum at r = √
. Next we note that
3
π
lim C(r) = lim C(r) = ∞
r→0+
r→∞
M ATH 150: S EMINAR #11 Q UESTIONS
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5
and conclude that C obtains the absolute minimum at r = √
.
3
π
5
The cost is minimized if the radius of the bottom of the pot is r = √
cm and if the height of the pot is
3
π
10
250
250
= √
h = 2 = 25
cm.
3
√
r π
π
3 2π
π
7. Each rectangular page of a book must contain 30 cm2 of printed text, and each page must have 2 cm
margins at top and bottom, and 1 cm margin at each side. What is the minimum possible area of such
a page?
Solution. Let x be the width of the page (in centimetres) and let y be the length of the page. The
question is to minimize the product A = xy under the constraint (x − 2)(y − 4) = 30.
We write the given constraint in the form
y =4+
30
x−2
and conclude that we need to minimize the function
30x
30
, x > 2.
= 4x +
A=x· 4+
x−2
x−2
From
dA
x−2−x
60
= 4 + 30 ·
=4−
dx
(x − 2)2
(x − 2)2
we conclude that
60
dA
=0⇔4=
⇔ (x − 2)2 = 15.
dx
(x − 2)2
√
Since x > 2, it follows that the only critical number is x = 2 + 15 ≈ 5.872983346 cm.
The fact that
d2 A
120
=
> 0, x > 2
dx2
(x − 2)3
implies that the function A = A(x) is concave upward
on its domain which means that it obtains the
√
absolute minimum at the critical number x = 2 + 15.
The minimum possible area of the page is
A(2 +
√
15) = 4 · (2 +
√
√
√
30 · (2 + 15)
√
15) +
= 38 + 8 15 ≈ 69
2 + 15 − 2
square centimetres.
8. You are supposed to design a rectangular box, open at the top and with a square base, that is to have
a volume of exactly 3500 cm3 . If the box is to require the least amount of material, what must be its
dimensions?
Solution. Let x ∈ (0, ∞) be the length of the side of the base of the box and let y be the height of the
box. Then
3500
Volume of the box = x2 y = 3500 ⇒ y = 2 .
x
The question is to find x (and y) so that the surface area of the box is minimal. Since the box is open at
the top its surface is given by
S = x2 + 4xy = x2 + 4x ·
3500
14000
= x2 +
.
x2
x
M ATH 150: S EMINAR #11 Q UESTIONS
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From
dS
x3 − 7000
14000
=
2
·
= 2x −
dx
x2
x2
it follows that
√
√
dS
3
3
= 0 ⇔ x3 − 7000 = 0 ⇔ x = 7000 = 10 7
dx
√
and we conclude that x = 10 3 7 is the only critical number. Since the function f (x) = x3 is increasing on
its domain we conclude that
√
dS
3
x > 10 7 ⇒ x3 > 7000 ⇒
>0
dx
and
√
dS
3
x < 10 7 ⇒ x3 < 7000 ⇒
< 0.
dx
By√the First Derivative Test, the function S = S(x) attains its local (and absolute) minimum at x =
10 3 7,
√
We conclude that the dimensions x = 10 3 7 cm and
y=
3500
35
3500
√
√ = √
=
cm.
3
(10 3 7)2
100 3 49
49
will give the least amount of material needed to construct the box.
9. A fence 8 ft tall runs parallel to a tall building at a distance of 4 ft from the building. What is the
length of the shortest ladder that will reach from the ground over the fence to the wall of the building?
Solution.
L
H
8
4
x
Figure 5: A building, a fence, and a ladder
Let x be the distance between the fence and the bottom of the ladder, and let H be the height of the top
of the ladder. Let L be the length of the ladder. See Figure 5. Observe that x > 0.
Then by the Pythagorean Theorem
L2 = H 2 + (x + 4)2 .
Observe two similar right triangles and conclude
H
x+4
4
=
⇒H =8 1+
.
8
x
x
Therefore, the question is to find x > 0 that minimizes the value of the function L = L(x) that is given
by
2
4
L2 = 64 1 +
+ (x + 4)2 .
x
From
4
4
256
2L(x)L (x) = 128 1 +
· − 2 + 2(x + 4) = 2(x + 4) 1 − 3
x
x
x
′
M ATH 150: S EMINAR #11 Q UESTIONS
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it follows that the only critical point of the function L = L(x) is given by
√
√
3
3
L′ (x) = 0 ⇔ x = 256 = 4 4.
Since x + 4 > 0 for x > 0 , it follows that
L′ (x) > 0 ⇔ 1 −
and
√
√
256
3
3
> 0 ⇔ x ∈ (4 4, ∞) ⇔ f is increasing on (4 4, ∞)
3
x
√
√
3
3
L′ (x) < 0 ⇔ x ∈ (0, 4 4) ⇔ f is decreasing on (0, 4 4).
√
Hence L(4 3 4) is a local and the absolute minimum value of the function L = L(x).
Finally, from
√
3
(L(4 4))2
2
√
4
3
64 1 + √
+ (4 4 + 4)2
3
4 4
√
√
64
3
3
= √
· (1 + 4)2 + 16(1 + 4)2 .
3
16
√
√
2
3
3
2
= 16(1 + 4) · √
+ 1 = 16(1 + 4)3 .
3
2
=
the length of the shortest ladder that will reach from the ground over the fence to the wall of the building
is
q
√
√
3
3
L = 4(1 + 4) 1 + 4.
10. A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent
into an equilateral triangle. How should the wire be cut so that the total area enclosed is a minimum?
Solution. Recall that the perimeter and the area of a square and a equilateral triangle are given by
the following table
Side Perimeter
Area
2
Square
a
P□ = 4a
A□ = a√
b2 3
Equilateral triangle
b
P△ = 3b A△ = 4
Suppose that the length of the piece of wire that is bent into the square is equal to L ∈ (0, 10). This
L
means that the perimeter of the square is P□ = L which implies that the side of the square is a = .
4
L2
.
Hence the area of the square is given by A□ =
4
Next we conclude that the length of the piece of wire that is bent into the triangle is equal to 10 − L.
This means that the perimeter of the triangle is P△ = 10 − L which implies
√ that the side of the triangle
2
10 − L
(10 − L) 3
is b =
. Hence the area of the triangle is given by A△ =
.
3
9·4
The question is to find the value of L ∈ (0, 10) that minimizes the function
√
L2
(10 − L)2 3
A(L) = A□ + A△ =
+
, L ∈ (0, 10).
16
36
To do so, we find the first derivative of the function A(L) to be able to determine its critical numbers:
√
dA
L (10 − L) 3
= −
.
dL
8
18
From
√
√
dA
L
(10 − L) 3
40 3
√
=0⇔ =
⇔L=
dL
8
18
9+4 3
M ATH 150: S EMINAR #11 Q UESTIONS
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√
40 3
√ . We observe that
we conclude that the only critical number is L =
9+4 3
√
d2 A
1
3
= +
>0
2
dL
8
18
√
40 3
√ minimizes the function
9+4 3
A(L) = A□ + A△ , L ∈ (0, 10). Therefore, by cutting the piece of wire into two pieces of the lengths
√
√
40 3
40 3
90
√ and 10 − L = 10 −
√ =
√
L=
9+4 3
9+4 3
9+4 3
for all L ∈ (0, 10). By the Second Derivative Test, the critical point L =
and then by bending the first piece into a square and the other into an equilateral will enclose the
minimal total area.
Note: One may consider the case that L ∈ [0, 10], i.e., the case that allows that the whole piece is bent
into an equilateral triangle (case L = 0) and that the whole piece is bent into a square (case L = 10.)
In this situation, since the given function as a polynomial is continuous, we use the Closed Interval
Method to obtain
√ !
40 3
√
≈ 2.72 < A(0) ≈ 4.8 < A(10) = 6.5.
A
9+4 3
11. Find an equation of the line through the point (3, 5) that cuts off the least area from the first quadrant.
Solution. Recall that all lines passing through the point (3, 5) are given by
y − 5 = m(x − 3).
Observe that the line that “cuts off ” a finite area from the first quadrant must intersect both the positive
ray of the x- and the positive ray of the y-axis. Hence, we are interested only in the case when m < 0.
(Sketch a graph if you need to convince yourself that this is true.)
Also observe that the area of the first quadrant that is cut off by the line y − 5 = m(x − 3), as the area
of the right triangle with the vertices (0, 0), (x0 , 0), and (0, y0 ), where x0 and y0 are the x− and the y−
intercepts of the line, is given by
x0 · y0
A=
.
2
Next we find the x- and the y intercepts:
0 − 5 = m(x0 − 3) ⇒ x0 =
3m − 5
m
and
y0 − 5 = m(0 − 3) ⇒ y0 = 5 − 3m.
It follows that our task is to find the value of m < 0 that minimizes the function
A = A(m) =
From
x0 · y0
(3m − 5)2
=−
.
2
2m
dA
1 6m(3m − 5) − (3m − 5)2
1 9m2 − 25
1
25
=− ·
=
−
·
=
−
·
9
−
,
dm
2
m2
2
m2
2
m2
and keeping in mind that m < 0, it follows that
dA
5
=0⇔m=− .
dm
3
M ATH 150: S EMINAR #11 Q UESTIONS
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Since
d2 A
1 50
=− · 3 >0
dm2
2 m
5
for all m < 0, by the Second Derivative Test, we conclude that m = − minimizes the function A = A(m).
3
Therefore, the line
5
y − 5 = − · (x − 3)
3
cuts off the least area from the first quadrant.
12. Use Newton’s method to approximate the negative root of ex = 4−x2 starting with initial approximation
x1 = −2 and finding x2 .
Solution. Let
f (x) = ex + x2 − 4
. Then f ′ (x) = ex + 2x and, for x1 = −2, by Newton’s method
x2
e−2 + 4 − 4
f (x1 )
= −2 −
′
f (x1 )
e−2 − 4
1
−2 + 8e2 − 1
8e2 − 3
= −2 −
=
=
2
2
1 − 4e
1 − 4e
1 − 4e2
≈ −1.96498.
=
x1 −
Note: Wolfram Alpha gives x ≈ −1.96464 as the negative solution of the equation ex = 4 − x2 .
√
13. Use Newton’s method to find the second approximation x2 of 5 31 starting with the initial approximation
x0 = 2.
√
Solution. The root of f (x) = x5 − 31 is 5 31, so we apply Newton’s method to f . We get that f ′ (x) = 5x4 ,
and Newton’s iterative formula is:
f (x0 )
25 − 31
1
159
=
2
−
=2−
=
f ′ (x0 )
5 · 24
80
80
f (xn )
f ′ (xn )
⇒
x2 = x1 −
159 5
− 31
f (x1 )
159
80
=
−
≈ 1.987340780.
159 4
f ′ (x1 )
80
5 · 80
xn+1 = xn −
We use x1 to find x2 :
x1 = x0 −
14. The equation
8x3 − 12x2 − 22x + 25 = 0
has a solution near x1 = 1. Use Newton’s method to find a better approximation x2 to this solution.
Express your answer as a fraction.
Solution. Our function is f (x) = 8x3 − 12x2 − 22x + 25 so f ′ (x) = 24x2 − 24x − 22. We get that
f (1) = 8 − 12 − 22 + 25 = −1 and f ′ (1) = 24 − 24 − 22 = −22 and so, from Newton’s iterative formula we
get:
f (x1 )
−1
1
21
f (xn )
⇒
x2 = x1 − ′
=1−
=1−
=
.
xn+1 = xn − ′
f (xn )
f (x1 )
−22
22
22
15. In this question we investigate the solution of the equation 2x − 1 = sin x.
(a) Explain why you know that the equation has at least one solution.
(b) Show that the equation has exactly one solutions.
(c) Use Newton’s method to approximate the solution of the equation starting with x1 = 0 and finding
x2 .
M ATH 150: S EMINAR #11 Q UESTIONS
S OLUTIONS
Solution.
(a) Let f (x) = 2x − 1 − sin x, whose roots would be solutions to our equation. This function f is
continuous, and we note that | sin x| ≤ 1, so f (100) > 0 and f (−100) < 0, which means by the
Intermediate Value Theorem, this function has at least one root, and thus this equation has at
least one solution.
(b) We note that f ′ (x) = 2 − cos x and | cos x| ≤ 1 so f ′ (x) > 0. Thus, this function is always increasing
and it has at most one root.
(c) We use f (0) = 0 − 1 − sin(0) = −1 and f ′ (0) = 2 − cos(0) = 1 in Newton’s Iterative Formula:
x2 = x1 −
−1
f (x1 )
=0−
= 1.
f ′ (x1 )
1