differentiation

304 DIFFERENTIAL CALCULUS
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
dy
dx = 3(−1.1)2 − 3,
which is positive.
If x is slightly more than −1, say −0.9, then
dy
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 dy
dx = 3x2 − 3, then d2 y
dx 2 = 6x
When x = 1, d2 y
is positive, hence (1, 3) is a
dx2 minimum value.
When x =−1, d2 y
is negative, hence (−1, 7)
dx2 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
dy
then
dθ = 4 − e−θ = 0
for a maximum or minimum value.
Hence 4 = e −θ ,
4 1 = eθ , giving θ = ln
4 1 =−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 = x3
3 − x2
2 − 6x + 5 and distinguish between
3
them. Sketch the graph.
Following the given procedure:
(i) Since y = x3
3 − x2
2 − 6x + 5 3 then
dy
dx = x2 − x − 6
(ii) At a turning point, dy = 0. Hence
dx
x 2 − x − 6 = 0, i.e. (x + 2)(x − 3) = 0,
from which x =−2orx = 3.
(iii) When x =−2,
y = (−2)3
3
When x = 3,
− (−2)2
2
− 6(−2) + 5 3 = 9
y = (3)3
3 − (3)2
2 − 6(3) + 5 3 =−115 6
Thus the co-ordinates ( of)
the turning points
are (−2, 9) and 3, −11
6
5 .
(iv) Since dy
dx = x2 − x − 6 then d2 y
dx 2 = 2x−1.
When x =−2,
d 2 y
= 2(−2) − 1 =−5,
dx2 which is negative.
Hence (−2, 9) is a maximum point.
When x = 3,
d 2 y
= 2(3) − 1 = 5,
dx2 which is positive.