Tutorial 8: Solutions
Applications of the Derivative
1. We are asked to find the absolute maximum and minimum values of f on the given
interval, and state where those values occur:
(a) f (x) = 2x3 + 3x2 − 12x on [1, 4].
The absolute extrema occur either at a critical point (stationary point or
point of non-differentiability) or at the endpoints. The stationary points are
given by f 0 (x) = 0, which in this case gives
6x2 + 6x − 12 = 0
=⇒
x = 1, x = −2.
Checking the values at the critical points (only x = 1 is in the interval [1, 4])
and endpoints:
f (1) = −7,
f (4) = 128.
Therefore the absolute maximum occurs at x = 4 and is given by f (4) = 124
and the absolute minimum occurs at x = 1 and is given by f (1) = −7.
(b) f (x) = (x2 + x)2/3 on [−2, 3]
As before, we find the critical points. The derivative is given by
f 0 (x) =
2 2x + 1
3 (x2 + x)1/2
and hence we have a stationary point when 2x + 1 = 0 and points of nondifferentiability whenever x2 + x = 0. Solving these, we get the three critical
points,
x = −1/2, and x = 0, −1.
Checking the value of the function at these critical points and the endpoints:
f (−2) = 22/3 ,
f (−1) = 0,
f (−1/2) = 2−4/3 ,
f (0) = 0,
f (3) = (12)2/3 .
Hence the absolute minimum occurs at either x = −1 or x = 0 since in both
these cases the minimum value is 0, while the absolute maximum occurs at
x = 3 and is f (3) = (12)2/3 .
(c) f (x) = x − 2 sin x on [−π/4, π/2]
The derivative is given by
f 0 (x) = 1 − 2 cos x
1
and the only critical point on [−π/4, π/2] is x = π/3. Checking the value of
the function at this critical point and at the end points:
√
f (−π/4) = −π/4 + 2 = 0.6288
√
f (π/3) = π/3 − 3 = −0.6849
f (π/2) = π/2 − 2 = −0.4292.
Hence the absolute minimum is at x = π/3 and is given by f (π/3) above
while the absolute maximum is at x = −π/4.
(d) f (x) = |6 − 4x| on [−3, 3]
The absolute value of a continuous differentiable function is everywhere differentiable except at the points where what’s inside the absolute value vanishes,
i.e., the only critical point is the point of non-differentiablility given by
6 − 4x = 0
=⇒
x = 3/2.
We now check the value of the function at this critical point and at the
endpoints:
f (−3) = 18, f (3) = 6, f (3/2) = 0.
Hence the absolute minimum occurs at x = 3/2 and is given by f (3/2) = 0
and the absolute maximum occurs at x = −3 and is given by f (−3) = 18.
(e) f (x) = x2 − x − 2 on (−∞, ∞)
A polynomial of even degree over the entire real line will have only one global
extremum which occurs at a stationary point, defined by
f 0 (x) = 2x − 1 = 0,
=⇒
x = 1/2.
It should be clear from the limiting behaviour as x → ±∞ that this point
has to be an absolute minimum. Alternatively, we could compute the second
derivative f 00 (x) = 2 > 0, which implies a minimum. So there is an absolute
minimum at x = 1/2 given by f (1/2) = −9/4. There is no absolute maximum.
(f) f (x) = x3 − 9x + 1 on (−∞, ∞)
This is a polynomial of odd degree and hence has no absolute extrema.
2. A closed rectangular container with a square base is to have a volume of 2000 cm3 .
It costs twice as much per square centimeter for the top and bottom as it does for
the sides. Find the dimensions of the container of least cost.
The volume is constrained by the equation
V = x2 h = 2000,
2
where x is the length of the side of the square base and h is the height. We are
trying to minimize the cost so we are looking for a function which represents the
total cost of manufacturing the container. Let p be the price/cm2 for the sides and
hence 2p is the price/cm2 for the top and bottom of the container. We have
cost of top and bottom = (area) × (price/cm2 )
= (2x2 ) × (2p) = 4x2 p
cost of sides = (area) × (price/cm2 )
= (4xh) × (p) = 4xhp.
Hence the total cost is
C = 4x2 p + 4xhp.
This is a function of both h and x so we use the constraint equation to eliminate
the h-dependence,
2000
h= 2
x
and hence the cost function is
C(x) = 4x2 p +
8000p
.
x
The absolute extrema on an open interval will occur at the critical points. The
stationary points are defined by
=⇒
8000p
=0
x2
x3 = 1000
=⇒
x = 10 cm.
C 0 (x) =8xp −
There is also a point of non-differentiability at x = 0 but this clearly isn’t in the
domain of interest. It is straighforward to show that f 00 (10) = 24p > 0 and hence
x = 10 cm is indeed the dimension of minimum cost. Subbing into the constraint
equation gives h = 20 cm.
3. The equation f (x) = x5 + x4 − 5 = 0 has one real solution. Use the Intermediate Value Theorem to obtain an initial estimate of the solution, then apply the
Newton-Raphson method to approximate the root to 5 decimal places.
Evaluating f (1) = −3 < 0 and f (2) = 43 > 0. Since the sign changes and
the function is continuous, the Intermediate Value Theorem tells us that there
must be a root in the interval (1, 2). Taking the midpoint of this interval to be our
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initial guess x1 = 1.5. Then the Newton-Raphson method gives
x2 = x1 −
x3 = x2 −
x4 = x3 −
x5 = x4 −
x6 = x5 −
f (x1 )
f 0 (x1 )
f (x2 )
f 0 (x2 )
f (x3 )
f 0 (x3 )
f (x4 )
f 0 (x4 )
f (x5 )
f 0 (x5 )
= 1.3027375201
= 1.232510136
= 1.22453239
= 1.224439563
= 1.22443955.
So to 5 decimal places, we have x = 1.22444.
4. Apply Rolle’s Theorem and use the fact that
d
(3x4 + x2 − 4x) = 12x3 + 2x − 4
dx
to show that the equation 12x3 +2x−4 = 0 has at least one solution in the interval
(0, 1).
We consider the function f (x) = 3x4 + x2 − 4x, which is continuous and differentiable on [0, 1] and which satisfies f (0) = 0, f (1) = 0. Hence Rolle’s Theorem
implies that there is at least one point x0 ∈ (0, 1) such that f 0 (x0 ) = 0, i.e., x0 is
a root of
12x3 + 2x − 4 = 0,
as required.
√
5. For the function f (x) = 25 − x2 , verify that the hypothesis of the Mean-Value
Theorem is satisfied on the interval [−5, 3] and find all values of c in that interval
that satisfy the conclusion of the theorem.
The hypothesis of the Mean Value Theorem requires that the function be continuous on some closed interval [a, b] and differentiable on the open interval (a, b). In
the present case, f (x) is continuous on the domain consisting of values satisfying
25 − x2 ≥ 0,
=⇒
Computing the derivative
f 0 (x) = − √
D(f ) = [−5, 5].
x
,
25 − x2
we see that the function is differentiable on the open interval (−5, 5). Hence
the hypothesis of the Mean Value Theorem is satisfied on the interval [−5, 3].
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The coordinates of the endpoints of this interval are A(−5, f (−5)) = (−5, 0) and
B(3, f (3)) = (3, 4) and the slope of this line-segent is
slope of AB =
4−0
1
= .
3 − (−5)
2
The Mean Value Theorem states that there is at least one value c such that f 0 (c)
is equal to the slope of this line segment, i.e.,
c
1
=
2
2
25 − c
2
c =5
√
c = ± 5.
−√
=⇒
=⇒
6. The function s(t) = t4 − 4t3 + 4t2 + 1, t ≥ 0, describes the position of a particle
moving along a coordinate line, where s is in metres and t is in seconds. Analyze
the motion of the particle by determining the intervals over which the particle
is moving in the positive or negative direction, the times when the particle has
stopped and the intervals over which the particle is speeding up or slowing down.
Give a schematic picture of the motion.
The velocity is given by
v(t) = s0 (t) = 4t3 − 12t2 + 8t = 4t(t − 1)(t − 2).
Hence the velocity is zero at t = 0, t = 1 and t = 2 which correspond to the
particle’s turning points. The sign of the velocity determines whether the particle
is moving in the positive or negative s-direction: The acceleration is
++++++++++++++++++++ 0
0
_ _ _ _ _ _ _ _ _ _ _ _ _ _
1
moving in pos. dir.
0 ++++++++++++++++++++
t
2
moving in neg. dir.
sign of v(t)
moving in pos. dir.
a(t) = 12t2 − 24t + 8 = 4(3t2 − 6t + 2)
which has roots at
1
t=1± √ .
3
The intervals on which the sign of the acceleration and velocity are the same are
the intervals over which the particle is speeding up and vice-versa.
5
+++++++++ 0
_ _ _ _ _ _ _ _ _ _ _ _ _ _
++++++++++++++++++++ 0
1
1− √
3
0
speeding up
0+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
_ _ _ _ _ _ _ _ _ _ _ _ _ _
0 ++++++++++++++++++++
1
1+ √
3
1
slowing down
speeding up
t
2
slowing down
sign of a(t)
sign of v(t)
speeding up
Finally, to draw the schematic diagram, we note the position of the particle
along the s-axis at each of these critical times:
√
s(0) = 1, s(1) = 2, s(2) = 1, s(1 ± 1/ 3) = 1.4444‘
. The schematic diagram is
1
t=1+ √
3
t=2
t=1
t=0
1
t=1− √
3
1
2
s
Figure 1: The red shaded regions are the intervals over which the particle is slowing
down
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