For a curve to concave upward, f ’’(x) should be positive. For a curve to concave downward,
f ’’(x) is –ve.
y  x2  x  2
y '  2x 1
y ''  2
1.
As 2nd derivative is +ve everywhere, this curve will be concave upward for all values of
x.
2. f ( x) 
x2 1
2x 1
f '( x) 
(2 x  1)2 x  2( x 2  1) 4 x 2  2 x  2 x 2  2 2 x 2  2 x  2


(2 x  1)2
(2 x  1)2
(2 x  1)2
f ''( x) 
(2 x  1)2 (4 x  2)  (2 x 2  2 x  2).4(2 x  1)
(2 x  1)4

2(2 x  1)3  8( x 2  x  1)(2 x  1)
(2 x  1)4
6(2 x  1)
(2 x  1) 4
6

(2 x  1)3

Hence, f ’’(x) will be +ve when
(2 x  1)3  0
Or,
2x 1  0
Or,
x
1
2
1
1
Hence, in the region, x   , the curve will be concave upward. For x   , the curve will be
2
2
concave downward.
3. f ( x) 
24
x  12
2
f '( x) 
48 x
( x  12)2
f ''( x) 
48( x 2  12)2  192 x 2 ( x 2  12)
( x 2  12)4

2
48( x 2  12)( x 2  12  4 x 2 )
( x 2  12)4
48(12  3x 2 )

( x 2  12)3
144( x 2  4)
 2
( x  12)3
We see that denominator will always be +ve. For f ‘’(x) to be +ve, x 2  4 or in the region x < -2
and x > 2, f ’’(x) will be +ve. Hence, in this region, the curve will be concave upward. In the
region, -2<x<2, the curve will be concave downward.
4.
y   x3  6 x 2  9 x  1
y '  3x 2  12 x  9
y ''  6 x  12
For y’’ to be +ve, -6x+12 > 0
Or, -6x > -12
Or, x > 2
Hence, the region, x> 2, the curve will be concave upward.
9. f ( x)  6 x  x 2
f '( x)  6  2 x
f ''( x)  2
Hence, f(x) will be maximum at 6 -2x = 0 or, x = 3; and maximum will be 18-9 = 9.
11. f ( x)  x3  5x 2  7 x
f '( x)  3x 2  10 x  7
f ''( x)  6 x  10
For f(x) to maximum or minimum, f ’(x) = 0. Hence,
3 x 2  10 x  7  0
Or,
3x 2  3x  7 x  7  0
Or,
3x(x-1)-7(x-1) = 0
Or,
(3x-1)(x-1) = 0
Or,
x = 1, 1/3
At these values of x, f ’’(x) is –ve; hence, relative maximum will occur at these values of x.
2
13.
f ( x)  x 3  3
f '( x) 
2  13
x
3
2 4
f ''( x)   x 3
9
We see that f ‘(x) will be zero at x = 0. Hence, relative extrema will occur at x = 0. At x = 0,
f ’’(x) is also zero; hence, we can not decide whether it is relative maximum or minimum.
2
15. f ( x)  x  1
f '( x) 
f ''( x) 
1

1 2
( x  1) 2 .2 x 
2
x
x2  1
1
3
( x 2  1) 2
Hence, relative extrema will occur at x = 0 as f’(x) is zero at x = 0 and it will be minimum as f’’(x) is +ve at
x = 0.
2
17. f ( x)  9  x
f '( x) 
x
9  x2
f ''( x) 
9
3
(9  x 2 ) 2
Relative extrema will occur at x = 0 and it will be maximum as f ‘’(x) is –ve at at x = 0.
16 x
8
19. f ( x)  x 2  2  f '( x)  ( x 2  2) 2
f ''( x) 
16( x 2  2)2  64 x 2 ( x 2  2)
48
 2
2
4
( x  2)
( x  2)2
Hence, relative extrema will occur at x = 0 and it will be minimum as f’’(x) is +ve at x = 0.
1
x
21. f ( x)  x  1  f '( x)  ( x  1) 2
f ''( x) 
2
( x  1)3
Hence, extrema will occur at x = ± ∞ and it will be maximum as f’’(x) is –ve at this value of x.
1
R

(600 x 2  x 3 )
61.
50000
R v/s x plot will be as shown below. R is on y-axis and x is on x-axis. We have diminishing returns for x >
600.
2000
1000
0
-1000 0
-2000
-3000
-4000
-5000
-6000
-7000
-8000
-9000
200
400
600
800
1000
1200
dR
1
2
Also dx  50000 (1200 x  3x )
d 2R
1

(1200  6 x)
2
dx
50000
R will have extrema when 1200 = 3x or x = 400. This can be sen from plot also. This will be maximumas
R’’(x) is –ve at x = 400.
3
2
63. N  0.12t  0.54t  8.22t
dN
 0.36t 2  1.08t  8.22
dt
d 2N
 0.72t  1.08
dt 2
We will have dN/dt maximum when N’’(t) is zero. Hence, at t = 1.08/0.72 = 1.5 or 8.30pm, N’ will be
maximum.
10000t 2
dx
(t 2  9)2t  t 2 .2t
18t
x



10000
 10000 2
2
2
2
67.
t 9
dt
(t  9)
(t  9)2
d 2x
9  5t
 18 10000 2
2
dt
(t  9)3
We will have x’ maximum when x’’ is zero. This will happen at t=9/5= 1.8.
The plot of x’ v/s t will be as shown below.
2500
2000
1500
1000
500
0
0
0.5
1
1.5
2
2.5
3
3.5
2.5
3
3.5
This is matching with the calculated values.
3
2
69. f ( x)  0.5 x  x  3x  5 It is blue line in plot
f '( x)  1.5x 2  2 x  3 It is red line in plot.
f ''( x)  3 x  2 It is green line in plot.
Plot will be as given below.
20
15
10
5
0
0
0.5
1
1.5
2
-5
In the range{0,3), we don’t have any extrema as curve is increasing everywhere.Same is the
behavior of f’ and f ‘’. We can not decide inflexion point from the plot in the given domain.
2
71. f ( x)  x 2  1 It is shown blue in plot.
f '( x) 
4 x
( x 2  1) 2 It is shown red in plot
f ''( x) 
12 x 2  4
( x 2  1)3 It is shown green in plot.
Plot will be as shown below.
3
2
1
0
-4
-3
-2
-1
0
1
2
3
4
-1
-2
-3
-4
-5
When f is increasing, f’’ is decreasing and when f is decreasing, f’’ is increasing. Inflexion point
is at x = 0.
x2
f
(
x
)

72.
x 2  1 It is shown blue in plot.
f '( x) 
2x
( x  1) 2 It is shown red in plot.
f '( x) 
2  10 x 2
( x 2  1)2 It is shown green in plot.
2
2.5
2
1.5
1
0.5
0
-4
-3
-2
-1
-0.5
0
1
2
3
4
-1
-1.5
-2
-2.5
Inflexion point is at x = 0. We have f minimum at x = 0. When f is increasing, f ‘’ is decreasing
and vice-versa.
2
73. C  0.5 x  10 x  7200
dC
 x  10
dx
For C to maximum,C’ = 0 or x+ 10 = 0or x = -10. We can not have –ve value of x. Hence C will be
minimum when there is no production.