Tutorial 7: SOLUTIONS
Analysis of Functions
1. For each of the following functions, we are asked to find (a) the intervals on which
f is increasing, (b) the intervals in which f is decreasing, (c) the open intervals on
which f is concave up, (d) the open intervals on which f is concave down, and (e)
the x-coordinates of all inflection points:
(a) f (x) = x2 − 3x + 8:
Differentiating gives
f 0 (x) = 2x − 3 > 0
for x > 3/2.
Hence f (x) is increasing on [−3/2, ∞) and decreasing on (−∞, 3/2].
The second derivative is f 00 (x) = 2 > 0 and hence the function is concave
up on the entire real line, which also implies that there are no inflection points.
(b) f (x) = x4 − 5x3 + 9x2 :
The derivative is given by
f 0 (x) = 4x3 − 15x2 + 18x = x(4x2 − 15x + 18).
Using the quadratic formula, it is straight-forward to show that the quadratic
factor in f 0 (x) has no real roots and hence the sign of the derivative changes
only at the point x = 0. The sign of f 0 (x) in each interval is determined by
subbing any value in a particular interval into the derivative. The following
schematic diagram represents the sign of the derivative:
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
+ + + + + + + + + + + + + + + + + + + sign of f ! (x)
0
Hence the function is increasing on [0, ∞)] and decreasing on (−∞, 0].
The second derivative is given by
f 00 (x) = 12x2 − 30x + 18 = 6(2x − 3)(x − 1).
The second derivative changes sign at x = 3/2 and x = 1 and again the sign
of f 00 (x) is determined by choosing any value in the interval. The schematic
diagram is given below. From the diagram, we see that the function is concave
up on (−∞, 1) and (3/2, ∞) and concave down on (1, 3/2), and the inflection
points are x = 1 and x = 3/2.
1
_ _ _ _ _ _ _ _ _ _ _ _ _
+++++++++++
+ + + + + + + + + + + sign of f !! (x)
3
2
1
x−2
:
− x + 1)2
The derivative is given by
(c) f (x) =
(x2
f 0 (x) =
−3(x2 − 3x + 1)
.
(x2 − x + 1)3
The stationary points correspond to the values where the derivative vanishes,
i.e., the roots of
√
3± 5
2
x − 3x + 1 = 0
=⇒
x=
.
2
There are also potential critical points of non-differentiability whenever the
denominator vanishes but in this case it is easy to see that the denominator
has no real roots and hence, the function has no points of non-differentiability.
√
Therefore, the sign of the derivative changes only at x = (3 ± 5)/2. The
schematic diagram is:
_ _ _ _ _ _ _ _ _ _ _ _ _
+++++++++++++++++
√
3− 5
2
_ _ _ _ _ _ _ _ _ _ _ _ _
sign of f ! (x)
√
3+ 5
2
√
√
√
Hence the function is increasing on [ 3−2 5 , 3+2 5 ] and decreasing on (−∞, 3−2 5 ]
√
and [ 3+2 5 , ∞).
The second derivative is given by
f 00 (x) =
6x(2x2 − 8x + 5)
.
(x2 − x + 1)4
The sign of the second derivative will change at the roots of the numerator
which are
2x2 − 8x + 5 = 0
√
4± 6
x = 0, x =
.
2
x=0
=⇒
and
The schematic diagram
is given below
from which we read off that f (x) is
√
√
4− 6
4+ 6
concave up on (0, 2 ) and ( 2 , ∞) and concave down on (−∞, 0) and
√
√
( 4−2 6 , 4+2 6 ). The inflection points are therefore x = 0 and x =
2
√
4± 6
2 .
_ _ _ _ _ _ _ +++++++++
_ _ _ _ _ _ _
sign of f !! (x)
√
4+ 6
2
√
4− 6
2
0
+++++++++
2. We are asked to show that x < tan x if 0 < x < π/2.
We start by defining the function f (x) = tan x − x on the interval [0, π/2). Clearly,
we have that f (0) = 0. The derivative of this function is
1
− 1.
cos2 x
f 0 (x) = sec2 x − 1 =
Since 0 < cos2 x < 1 on (0, π/2), we have that
f 0 (x) > 0
on
(0, π/2),
and hence we say that f (x) is increasing on [0, π/2). We already know that the
function evaluates to zero at x = 0 and since it is increasing it must evaluate to
something greater than zero for all x ∈ (0, π/2). Hence
f (x) > 0
on
(0, π/2)
=⇒
x < tan x
on
(0, π/2).
3. For each of the following functions, we are asked to locate the critical points and
identify which critical points are stationary points:
(a) f (x) = 4x4 − 16x2 + 17.
This is a polynomial which is everywhere differentiable and hence the only
critical points are stationary points, which are defined by vanishing derivative:
f 0 (x) = 16x3 − 16x = 16x(x − 1)(x + 1) = 0
=⇒
x = 0, x = 1, x = −1
are the critical points.
x2
.
+8
The derivative is given by
(b) f (x) =
x3
f 0 (x) =
x(16 − x3
.
(x3 + 8)2
The stationary points are given by the roots of the numerator
√
3
x(16 − x3 ) = 0
=⇒
x = 0, x = (16)1/3 = 2 2.
3
The points of non-differentiability are given by the roots of the denominator,
x3 + 8 = 0
=⇒
x = −2.
(c) f (x) = | sin x|.
To compute the derivative, we take u = sin x, and then f (x) = |u|. Hence
f 0 (x) =
df
df du
d|u|
=
= cos x
.
dx
du dx
du
The stationary points are defined by vanishing derivative, i.e., whenever
cos x = 0
=⇒
x = ±π(n + 21 ),
n = 0, 1, 2...
We also saw in class that the absolute value function |u| is not differentiable
at u = 0, and hence in the present case, the points of non-differentiability are
defined by
sin x = 0
x = ±nπ,
=⇒
n = 0, 1, 2....
4. For the polynomial p(x) = x(x2 − 1)2 , we are asked to find the following and hence
sketch the graph:
(a) The coordinates of the x and y-intercepts.
The graph cuts the y-axis at x = 0, which gives y = 0, i.e. at the origin (0, 0).
The graph cuts the x-axis at y = 0, which gives x = 0, x = ±1. The roots
x = ±1 are roots of multiplicity 2, and hence we would expect the x-axis to
be tangent to the graph at these points.
(b) The stationary points.
The stationary points are defined by vanishing derivative, and one obtains
f 0 (x) = (x2 − 1)(5x2 − 1) = 0
=⇒
1
x = ±1, x = ± √ .
5
(c) The intervals over which f is increasing and decreasing.
The intervals of increase/decrease are determined by the sign of the derivative,
which changes only when x passes through one of the stationary points above.
Plugging in any value in an interval determines the sign of the derivative in
that interval. This is represented in the schematic below, from which we
conclude that f (x) is increasing on (−∞, −1], [− √15 , √15 ] and [1, ∞) and is
decreasing on [−1, − √15 ] and [ √15 , 1].
4
++++++
_ _ _ _ _
++++++
1
√
5
1
−√
5
−1
_ _ _ _ _
++++++
sign of f ! (x)
1
(d) The intervals over which f is concave up/down.
The intervals of concavity are determined by the sign of the second derivative,
which is
f 00 (x) = 4x(5x2 − 3),
q
which changes sign at x = 0, and x = ± 35 . This is represented in the
following schematic:
_ _ _ _ _ _ _
−
!
+++++++++
3
5
_ _ _ _ _ _ _
!
0
+++++++++
sign of f !! (x)
3
5
q
q
Hence f (x) is concave up on (− 35 , 0) and ( 35 , ∞), and concave down on
q
q
(−∞, − 35 ) and (0, 35 ).
(e) Inflection points.
These are the points where the functions changes concavity,
q and therefore in
the present case, the inflection points are x = 0, x = ± 35 .
A sketch of the graph is given below:
5
5. For the rational function f (x) = x2 /(x2 − 4), we are asked to find the following
and hence sketch the graph:
(a) The symmetries of the function.
Since the expression for f (x) only contains even powers of x, we have that
f (−x) = f (x) and hence the graph will be symmetric about the y-axis.
(b) The coordinates of the x and y-intercepts.
The graph cuts the y-axis at x = 0 which gives y = 0, i.e., the graph passes
through the origin. The graph cuts the x-axis at y = 0, which again just gives
x = 0, so the only intercept is the origin. We also note that x = 0 is clearly
a root of multiplicity 2.
(c) The horizontal and vertical asymptotes.
The horizontal asymptotes are defined by the limit
1
= 1,
x→±∞ (1 − 4/x2 )
lim f (x) = lim
x→±∞
which implies y = 1 is the horizontal asymptote.
The vertical asymptotes are defined by vanishing denominator,
x2 − 4 = 0
=⇒
x = ±2.
(d) The stationary points.
The stationary points are the points where the derivative is zero,
f 0 (x) = −
(x2
8x
=0
− 4)2
=⇒
x = 0.
We note however that this is not the only critical point since the function is
discontinuous at x = ±2 and hence not differentiable at x = ±2.
(e) The intervals over which f is positive and negative.
The only points at which the sign of the function can possibly change are at
the roots or at discontinuities. The only root in this case has multiplicity 2 and
so does not cross the x-axis so the sign of the function will not change as we
pass through x = 0. The sign of the function will change as we pass through
the discontinuities however. This is represented by the schematic diagram
below, where as per usual the sign in any interval is simply determined by
plugging in any value in that interval. From the diagram, we read off that
f (x) is positive on (−∞, −2) and (2, ∞) and negative on (−2, 2).
6
+++++++++
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _
0
−2
+++++++++
sign of f (x)
2
(f) The intervals of increase/decrease.
These intervals are determined by the sign of the derivative. From the expression for f 0 (x) above, the denominator is clearly always positive and hence the
sign is determined only by the numerator which changes sign only at x = 0.
We note also that the function is not differentiable at x = ±2, although the
sign of the derivative does not change sign as we pass through these critical
points. This is represented in the schematic diagram below:
+++++++++
∞++++++++
−2
_ _ _ _ _ _ _
0
∞
_ _ _ _ _ _ _
sign of f ! (x)
2
Hence f (x) is increasing on (−∞, −2) and (−2, 0] and decreasing on [0, 2) and
(2, ∞), where we note that we have excluded the points of non-differentiablility
from the intervals. (These are also discontinuities of the function and hence
don’t lie in the domain of f (x)).
(g) The intervals of concavity.
These are determined by the sign of the second derivative where
f 00 (x) =
8(3x2 + 4)
.
(x2 − 4)3
It is clear that the numerator is always positive and hence the sign of f 00 (x)
is determined by the sign of the denominator, which is negative whenever
x2 − 4 < 0
=⇒
−2 < x < 2.
Hence f (x) is concave up on (−∞, −2) and (2, ∞) and concave down on
(−2, 2).
(h) Inflection points.
The inflection points are where the concavity changes, which in this case is
x = ±2. It is worth pointing out that this is a case where the inflection points
are not the points where f 00 (x) = 0.
A sketch of the graph is given below:
7
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