Some applications of differentiation 305
(a) Considering the point (1, 3):
If x is slightly less than 1, say 0.9, then
d y
dx
= 3(0.9)
2
− 3,
which is negative.
If x is slightly more than 1, say 1.1, then
d y
dx
= 3(1.1)
2
− 3,
which is positive.
Since the gradient changes from negative to positive, the point (1, 3) is a minimum point.
Considering the point (−1, 7):
If x is slightly less than −1, say −1.1, then
d y
dx
= 3(−1.1)
2
− 3,
which is positive.
If x is slightly more than −1, say −0.9, then
d y
dx
= 3(−0.9)
2
− 3,
which is negative.
Since the gradient changes from positive to negative, the point (−1, 7) is a maximum point.
(b) Since
d y
dx
= 3x 2 − 3, then
d 2 y
dx 2 = 6x
When x = 1,
d 2 y
dx 2 is positive, hence (1, 3) is a
minimum value.
When x = −1,
d
2 y
dx 2 is negative, hence (−1, 7) is
a maximum value.
Thus the maximum value is 7 and the minimum value is 3.
It can be seen that the second differential method of
determining the nature of the turning points is, in
this case, quicker than investigating the gradient.
Problem 12. Locate the turning point on the
following curve and determine whether it is a
maximum or minimum point: y = 4θ + e −θ .
Since
y = 4θ + e −θ
then
d y
dθ
= 4 − e −θ = 0
for a maximum or minimum value.
Hence 4 = e
−θ ,
1
4 = e
θ , giving θ = ln
1
4 =−1.3863 (see
Chapter 4).
When θ = − 1.3863, y = 4(−1.3863) + e −(−1.3863)
= 5.5452 +4.0000 =−1.5452
Thus (−1.3863, −1.5452) are the co-ordinates of the
turning point.
d 2 y
dθ 2 = e
−θ
.
When θ =−1.3863,
d 2 y
dθ 2 = e
+1.3863
= 4.0,
which is positive, hence (−1.3863, −1.5452) is a
minimum point.
Problem 13. Determine the co-ordinates of the
maximum and minimum values of the graph
y =
x 3
3
−
x 2
2
− 6x +
5
3
and distinguish between
them. Sketch the graph.
Following the given procedure:
(i) Since y =
x 3
3
−
x 2
2
− 6x +
5
3
then
d y
dx
= x 2 − x − 6
(ii) At a turning point,
d y
dx
= 0. Hence
x 2 − x − 6 = 0, i.e. (x + 2)(x − 3) = 0,
from which x = −2 or x = 3.
(iii) When x =−2,
y =
(−2) 3
3
−
(−2) 2
2
− 6(−2) +
5
3
= 9
When x = 3,
y =
(3)
3
3
−
(3)
2
2
− 6(3) +
5
3
= −11
5
6
Thus the co-ordinates of the turning points are
(−2, 9) and
3, −11
5
6
.
(iv) Since
d y
dx
= x
2
− x − 6 then
d 2 y
dx 2 = 2x−1.
When x =−2,
d 2 y
dx 2 = 2(−2) − 1 = −5,
which is negative.
(a) Considering the point (1, 3):
If x is slightly less than 1, say 0.9, then
d y
dx
= 3(0.9)
2
− 3,
which is negative.
If x is slightly more than 1, say 1.1, then
d y
dx
= 3(1.1)
2
− 3,
which is positive.
Since the gradient changes from negative to positive, the point (1, 3) is a minimum point.
Considering the point (−1, 7):
If x is slightly less than −1, say −1.1, then
d y
dx
= 3(−1.1)
2
− 3,
which is positive.
If x is slightly more than −1, say −0.9, then
d y
dx
= 3(−0.9)
2
− 3,
which is negative.
Since the gradient changes from positive to negative, the point (−1, 7) is a maximum point.
(b) Since
d y
dx
= 3x 2 − 3, then
d 2 y
dx 2 = 6x
When x = 1,
d 2 y
dx 2 is positive, hence (1, 3) is a
minimum value.
When x = −1,
d
2 y
dx 2 is negative, hence (−1, 7) is
a maximum value.
Thus the maximum value is 7 and the minimum value is 3.
It can be seen that the second differential method of
determining the nature of the turning points is, in
this case, quicker than investigating the gradient.
Problem 12. Locate the turning point on the
following curve and determine whether it is a
maximum or minimum point: y = 4θ + e −θ .
Since
y = 4θ + e −θ
then
d y
dθ
= 4 − e −θ = 0
for a maximum or minimum value.
Hence 4 = e
−θ ,
1
4 = e
θ , giving θ = ln
1
4 =−1.3863 (see
Chapter 4).
When θ = − 1.3863, y = 4(−1.3863) + e −(−1.3863)
= 5.5452 +4.0000 =−1.5452
Thus (−1.3863, −1.5452) are the co-ordinates of the
turning point.
d 2 y
dθ 2 = e
−θ
.
When θ =−1.3863,
d 2 y
dθ 2 = e
+1.3863
= 4.0,
which is positive, hence (−1.3863, −1.5452) is a
minimum point.
Problem 13. Determine the co-ordinates of the
maximum and minimum values of the graph
y =
x 3
3
−
x 2
2
− 6x +
5
3
and distinguish between
them. Sketch the graph.
Following the given procedure:
(i) Since y =
x 3
3
−
x 2
2
− 6x +
5
3
then
d y
dx
= x 2 − x − 6
(ii) At a turning point,
d y
dx
= 0. Hence
x 2 − x − 6 = 0, i.e. (x + 2)(x − 3) = 0,
from which x = −2 or x = 3.
(iii) When x =−2,
y =
(−2) 3
3
−
(−2) 2
2
− 6(−2) +
5
3
= 9
When x = 3,
y =
(3)
3
3
−
(3)
2
2
− 6(3) +
5
3
= −11
5
6
Thus the co-ordinates of the turning points are
(−2, 9) and
3, −11
5
6
.
(iv) Since
d y
dx
= x
2
− x − 6 then
d 2 y
dx 2 = 2x−1.
When x =−2,
d 2 y
dx 2 = 2(−2) − 1 = −5,
which is negative.