Arvind Borde: Math 7
Homework 8, Part a, solutions, page 192, 3–14
and f 00 (x) < 0 when
4x2
< 1 or 4x2 < x2 + 12 or 3x2 < 12 or x2 < 4.
x2 + 12
f (x) is concave up (f 00 (x) > 0) on (−∞, −2) and (2, ∞).
f (x) is concave down (f 00 (x) < 0) on (−2, 2).
8.
f 0 (x) =
=
f 00 (x) =
=
NOTE: The graphs below are graphs of the original functions.
Red lines are concave up and blue lines are concave down.
3. y 0 = 2x − 1, y 00 = 2 > 0.
Graph of y is concave up throughout.
4. g 0 (x) = 6x − 3x2 , g 00 (x) = 6 − 6x = 6(1 − x).
g(x) is concave up (g 00 (x) > 0) when 1 > x.
g(x) is concave down (g 00 (x) < 0) when 1 < x.
5. f 0 (x) = −3x2 + 12x − 9, f 00 (x) = −6x + 12 = 6(−x + 2).
f (x) is concave up (f 00 (x) > 0) when x < 2.
f (x) is concave down (f 00 (x) < 0) when x > 2.
=
12x3 + 4x − 12x3
4x(3x2 + 1) − 2x2 (6x)
=
2
2
(3x + 1)
(3x2 + 1)2
4x
.
2
(3x + 1)2
4(3x2 + 1)2 − 4x[2(3x2 + 1)(6x)]
(3x2 + 1)4
(3x2 + 1)[4(3x2 + 1) − (8x)(6x)
(3x2 + 1)4
12x2 + 4 − 48x2
4 − 36x2
=
.
2
3
(3x + 1)
(3x2 + 1)3
Therefore, f 00 (x) > 0 when
4 > 36x2 , or x2 < 4/36 = 1/9, or − 1/3 < x < 1/3,
and f 00 (x) < 0 when
4 < 36x2 , or x2 > 4/36 = 1/9, or x < −1/3, x > 1/3.
f (x) is concave up (f 00 (x) > 0) on (−1/3, 1/3).
f (x) is concave down (f 00 (x) < 0) on (−∞, −1/3) and (1/3, ∞).
9.
f 0 (x) =
=
6. h0 (x) = 5x4 − 5, h00 (x) = 20x3 .
h(x) is Concave up (h00 (x) > 0) when x > 0.
h(x) is concave down (h00 (x) < 0) when x < 0.
f 00 (x) =
=
=
7.
2
f (x) = 24(x + 12)
0
2
−1
f (x) = −24(x + 12)
.
−2
(2x) = −48x(x2 + 12)−2 .
f 00 (x) = −48(x2 + 12)−2 − 48x[−2(x2 + 12)−3 (2x)]
= −48(x2 + 12)−2 + 48 × 4x2 [(x2 + 12)−3 ]
= −48(x2 + 12)−2 [−1 + 4x2 (x2 + 12)−1 ]
−48
4x2
= 2
−1 + 2
(x + 12)2
x + 12
Therefore, f 00 (x) > 0 when
4x2
> 1 or 4x2 > x2 + 12
x2 + 12
or
3x2 > 12
or x2 > 4,
2x(x2 − 1) − (x2 + 1)(2x)
(x2 − 1)2
3
2x − 2x − 2x3 − 2x
−4x
= 2
(x2 − 1)2
(x − 1)2
−4(x2 − 1)2 + 4x[2(x2 − 1)(2x)]
(x2 − 1)4
2
4(x − 1)[−(x2 − 1) + 4x2 ]
(x2 − 1)4
2
4[−x + 1 + 4x2 ]
4(3x2 + 1)
=
.
(x2 − 1)3
(x2 − 1)3
The numerator here is always positive. The sign of f 00 (x) will
depend on the sign of (x2 − 1):
f 00 (x) > 0 when x2 > 1 and f 00 (x) < 0 when x2 < 1.
f (x) is concave up (f 00 (x) > 0) on (−∞, −1) and (1, ∞).
f (x) is concave down (f 00 (x) < 0) on (−1, 1).
12.
2x(2x − 1) − (x2 − 1)(2)
(2x − 1)2
2x2 − 2x + 2
4x2 − 2x − 2x2 + 2
=
.
=
2
(2x − 1)
(2x − 1)2
(4x − 2)(2x − 1)2 − (2x2 − 2x + 2)[2(2x − 1)(2)]
h00 (x) =
(2x − 1)4
(4x − 2)(2x − 1)2 − 4(2x2 − 2x + 2)(2x − 1)
=
(2x − 1)4
(2x − 1)[(4x − 2)(2x − 1) − 4(2x2 − 2x + 2)]
=
(2x − 1)4
2
8x − 4x − 4x + 2 − 8x2 + 8x − 8
−6
=
=
(2x − 1)3
(2x − 1)3
The numerator here is always negative. The sign of h00 (x) will
depend on the sign of (2x − 1):
h00 (x) > 0 when 2x < 1 and h00 (x) < 0 when 2x > 1.
h(x) is concave up (h00 (x) > 0) on (−∞, 1/2).
h(x) is concave down (h00 (x) < 0) on (1/2, ∞).
h0 (x) =
1
1
(−3x5 + 40x3 + 135x)0 =
(−15x4 + 120x2 + 135).
10. y 0 =
270
270
1
1
2x
(−60x3 + 240x) =
(60x)(−x2 + 4) =
(−x2 + 4).
y 00 =
270
270
9
y 00 > 0 when x and (−x2 + 4) have the same sign.
y 00 < 0 when x and (−x2 + 4) have opposite signs.
It is easy to check to see the following:
x < −2: y 00 > 0, therefore y is concave up.
−2 < x < 0: y 00 < 0, therefore y is concave down.
0 < x < 2: y 00 > 0, therefore y is concave up.
2 < x: y 00 < 0, therefore y is concave down.
11.
g 0 (x) =
=
g 00 (x) =
=
=
2x(4 − x2 ) − (x2 + 4)(−2x)
(4 − x2 )2
3
8x − 2x + 2x3 + 8x
16x
=
.
2
2
(4 − x )
(4 − x2 )2
16(4 − x2 )2 − 16x[2(4 − x2 )(−2x)]
(4 − x2 )4
2 2
16(4 − x ) + 64x2 (4 − x2 )
(4 − x2 )4
2
16(4 − x )[(4 − x2 ) + 4x2 ]
16(4 + 3x2 )
=
(4 − x2 )4
(4 − x2 )3
The numerator here is always positive. The sign of g 00 (x) will
depend on the sign of (4 − x2 ):
g 00 (x) > 0 when x2 < 4 and g 00 (x) < 0 when x2 > 4.
g(x) is concave up (g 00 (x) > 0) on (−2, 2).
g(x) is concave down (g 00 (x) < 0) on (−∞, −2) and (2, ∞).
13. y 0 = 2 − sec2 x, y 00 = −[2 sec x][sec x tan x] = 2 sec2 x tan x.
Since sec2 x > 0, the sign of y 00 will depend on the sign of tanx
on the given interval (−π/2, π/2). The graph of tan x shows that
it is positive in this interval for x > 0 and negative for x < 0.
Therefore,
y is concave up (y 00 > 0) on (−π/2, 0), and
y is concave down (y 00 < 0) on (0, π/2).
14. y = x + 2(sin x)−1 , so y 0 = 1 − 2(sin x)−2 cos x.
4 cos x
2
y 00 = 4(sin x)−3 cos x − 2(sin x)−2 (− sin x) =
+
3
sin x
sin x
If you plot the answer for y 00 , you will see that it is positive when
x > 0 and negative when x < 0. So,
y is concave up (y 00 > 0) on (0, π), and
y is concave down (y 00 < 0) on (−π, 0).