Mathematical Techniques A: Solutions Exercise Set 3
October 24, 2015
Exercise 3
Higher order derivatives and identifying maxima and minima of a function
1. By differentiation find the turning point of the function y = 2x2 + 12x + 1, and determine
whether it is a maximum or a minimum. Sketch the graph of the function.
2. By differentiation find the turning point of the function y = x2 + x − 6, and determine whether
it is a maximum or a minimum. Sketch the graph of the function.
3. Show by differentiation that the graphs of y = x2 + 6x + 9 and y = −x2 − 6x − 9 are both
tangent to the x - axis at x = −3, but that the first of these functions reaches a minimum at this
point, and the second a maximum. Confirm your result by solving the equations x2 + 6x + 9 = 0
and −x2 − 6x − 9 = 0. Sketch the graphs of the functions.
4. Show by differentiation that the function y = x3 + 2x2 + x + 9 has a maximum at x = −1, a
minimum at x = −1/3, and a point of inflexion at x = −2/3. Sketch the function.
5. By examining the signs of the first and second derivatives, determine the slope and curvature
of the function y = −x3 + 9x at the point x = 4. Hence draw a rough sketch graph of the curve in
the vicinity of this point.
6. By examining the signs of the first and second derivatives, determine the slope and curvature
of the function y = x2 + 5x + 3 for values of x lying between 0 and 1. (A range of values such as
this is called an interval). Hence draw a rough sketch graph of the curve in the interval 0 < x < 1.
7. Given the function y = 1/x , examine the signs of the first and second derivatives and hence
show
(i)
(ii)
(iii)
that the function is negatively sloped for all x 6= 0;
that the function is convex when x > 0 and concave when x < 0.
Use this information to sketch the graph of the function.
Mathematical Techniques A
2
Solutions
1. In order to find the turning point and determine if its a maximum or a minimum, we need to
find the first and second derivative of the function and the point at which the first derivative is equal
to zero along with the sign of the second derivative at that point:
dy
= 0 ⇒ 4x + 12 = 0 ⇒ x = −3
dx
d2 y
= 4>0
dx2
Given that the second derivative is always positive our function is convex and we have a minimum
at x = −3. At that point the value of the function is y = 18 − 36 + 1 = −17. The graph of the
function is depicted below.
y
y = 2x2 + 12x + 1
40
20
x
−8
−6
−4
−2
2
−20
2. Again in order to find the turning point and determine if its a maximum or a minimum, we need
to find the first and second derivative of the function and the point at which the first derivative is
equal to zero along with the sign of the second derivative at that point:
dy
= 0 ⇒ 2x + 1 = 0 ⇒ x = −1/2
dx
d2 y
= 2>0
dx2
Given that the second derivative is always positive our function is convex and we have a minimum
at x = −1/2. At that point the value of the function is y = 0.25 − 0.5 − 6 = −6.25. The graph of
the function is depicted below.
y
y = x2 + x − 6
40
20
x
−8
−6
−4
−2
2
4
6
−20
EC121
Omiros Kouvavas
Mathematical Techniques A
3
3. Once in order to find the turning point and determine if its a maximum or a minimum, we need
to find the first and second derivative of the functions and the point at which the first derivative is
equal to zero along with the sign of the second derivative at that point.Fort the first one we have:
dy
dx
d2 y
dx2
=
0 ⇒ 2x + 6 = 0 ⇒ x = −3
=
6>0
Given that the second derivative is always positive our function is convex and we have a minimum
at x = −3. At that point the value of the function is y = 9 − 18 + 9 = 0.
For the second one we have:
dy
dx
d2 y
dx2
0 ⇒ −2x − 6 = 0 ⇒ x = −3
=
= −6 < 0
Given that the second derivative is always negative our function is concave and we have a maximum
at x = −3. At that point the value of the function is y = −9 + 18 − 9 = 0. The graph of the two
functions is depicted below.
y
y = x2 + 6x + 9
y = −x2 − 6x − 9
20
x
−8
−6
−4
−2
2
−20
4. Once more, in order to find the turning point and determine if its a maximum or a minimum,
we need to find the first and second derivative of the function and the points at which the first
derivative is equal to zero along with the sign of the second derivative at each point.Furthermore we
need to find the point where the second derivative equal zero to identify the inflection point:
dy
dx
=
0 ⇒ 3x2 + 4x + 1 = 0
(1)
Now using the quadratic formula we have:
x1,2 =
−b ±
√
b2 − 4ac
2a
=
−4 ±
6
√
4
(2)
EC121
Omiros Kouvavas
Mathematical Techniques A
4
Thus x1 = − 13 and x2 = −1. Now that we found the two points the first derivative is equal to zero
we need to check the sign of the second derivative in each of this points:
d2 y
dx2
2
d y
|x
dx2 1
d2 y
|x
dx2 2
=
6x + 4
=
2>0
= −2 < 0
(3)
Thus for x1 = − 31 we have a minimum and for x2 = −1 we have a maximum. Finally, for x = − 32
the
d2 y
dx2
= 0 and we have an inflection point. Below, we can see the graph of the function.
100 y
50
x
−4
−2
2
4
−50
−100
5.
To determine the slope we need to find and evaluate the first derivative at x = 4:
dy
|4
dx
d2 y
|4
dx2
=
−39 < 0
=
−24 < 0
Given that the first derivative is negative at x = 4 the function is negatively sloped. Furthermore,
the second derivative is also negative at x = 4, which means the slope is becoming more negative
you can observe that in the graph below.
20
y
x
−4
−2
2
4
−20
−40
−60
−80
EC121
Omiros Kouvavas
Mathematical Techniques A
5
First we can analyze the sign of f (x) given x = ±3 when f (x) = 0:
f (x)
−∞
+
+
−3
0
−
√
− 3
−
√
0
0
−
3
+
+
+
3
0
+∞
−
−
dy
We can analyze the slope of the function using the sign of the first derivative ( dx
= −3x2 + 9 where
√
dy
x = ± 3 when dx = 0) as in the table below:
f 0 (x)
f (x)
−∞
−
&
−
&
−3
−
&
√
− 3
0
−
&
+
%
√
0
+
%
3
0
+
%
−
&
3
−
&
−
&
+∞
−
&
2
d y
We can also analyze the curvature of the function using the sign of the second derivative ( dx
2 = −6x
where its positive for negative values of x and negative otherwise) as in the table below:
f 00 (x)
f (x)
6.
−∞
+
∪
+
∪
−3
+
∪
+
∪
√
− 3
+
∪
0
0
+
∪
√
−
∩
3
−
∩
−
∩
3
−
∩
−
∩
+∞
−
∩
First we need to find the two derivatives:
dy
dx
d2 y
dx2
=
2x + 5
=
2>0
The first derivative is positive for x > 0 and negative otherwise. The second derivative is always
positive. Given this information we can determine that the function is downward sloping up to zero
and upward sloping from then on. Furthermore, the function is always convex. For the interval
between zero and 1 the function is only upward sloping as the first derivative is strictly positive for
values of x ∈ (0, 1). In order to draw the graph for the interval x ∈ (0, 1) we need to find f (0),f (1)
and join them with a convex upward sloping function.
f (0)
=
3
f (1)
=
1+5+3=9
So now we can draw the graph:
EC121
Omiros Kouvavas
Mathematical Techniques A
6
y
8
6
4
2
x
0.2
0.4
0.6
0.8
1
We can also graph the function for a larger interval:
y
60
40
20
x
−8
−6
−4
−2
2
4
6
7.
(i)
First of all the function is not defined at zero. Taking the first derivative we have1
dy
1
= − 2 < 0 ∀x ∈ (−∞, 0) ∪ (0, +∞)
dx
x
given we have determined the slope as negative the function is downward sloping in all the points it
is defined(The function is not defined when x = 0).
(ii)
Taking the second derivative:
d2 y
= 2x−3 < 0 ∀x ∈ (−∞, 0) ,
dx2
d2 y
> 0 ∀x ∈ (0, +∞)
dx2
(iii) Until now we know that our function is not defined at zero, is negatively slope in both
sides of the x-axis and is concave up to zero and convex afterwards. Observing the functional form
(or plugging some values) we can also determine that the function is asymptotic to the y = 0 from
below as x → −∞,and asymptotic to −∞ as x → 0 from negative values. Similarly the function
is asymptotic towards +∞ as x → 0 from positive values and to zero as x → +∞. Using, all this
information and combining it with the slope and curvature we can graph our functions as below.
1 Same
EC121
mathematical symbol explanation: ∀:for all,∈:in,∪:union
Omiros Kouvavas
Mathematical Techniques A
7
10
F (x)
1/x
5
x
−4
−2
2
4
−5
−10
EC121
Omiros Kouvavas