Problem 1 :
Find intervals of concavity and points of inflexion for the following functions:
(i) f (x) = x(x − 4)3
(ii) y = sin x + cos x, 0 < x < 2π
f(x) = 1/2 (ex-e-x)
Problem 2 :
Find the local extrema for the following functions using second derivative test :
(i) f(x) = −3x5+ 5x3
(ii) f(x) = x logx
(iii) f(x) = x2e-2x Solution
Problem 3 :
For the function
f(x) = 4x3+3x2-6x+1
find the intervals of
(i) monotonicity
(ii) local extrema
(iii) intervals of concavity and
(iv) points of inflection. Solution
Problem 4 :
What are the coordinates of the points of inflection for the graph of f(x) = sin2x on the closed interval [0, π] ?
a) x = 3π/4 only
b) x = π/4, x = π/2 and x = 3π/4
c) x = π/4 and x = 3π/4
d) x = π/2 e) x = π/4
Problem 5 :
The function g is defined by the equation
g(x) = 6x5 - 10x3
Determine the values of x for which the graph of function g is concaved upwards
a) x > 1/2 b) -√2/2 < x < 0 or x > √2/2
c) -1/2 < x < 0 or x > 1/2 d) -1/2 < x < 1/2
e) -√2/2 < x < √2/2
Application of first derivatives
Answers :
Problem 1 :
(i) Concave up on (-∞, 2) and (4, π).
Concave down on (2, 4).
point of inflection are (2, -16) and (4, 0).
(ii) Concave down on (0, 3π/4) and (7π/4, 2π).
Concave up on (3π/4, 7π/4).
Point of inflection are (3π/4, 0) and (7π/4, 0).
(iii) Concave up on (-∞, 0) and concave down on (0, ∞)
point of inflection is (0, 0).
Problem 2 :
(i) local maximum point is (1, 2) and local maximum is 2.
(ii) local minimum is -1/e
(iii) Local maximum = 1/e2 and Local minimum = 0
Problem 3 :
Concave downward on (-∞, -1/4) and Concave upward on (-1/4, ∞). Point of inflection is (-1/4, 21/8)
Problem 4 :
So, in the interval [0, π] the given function has point of inflection at the point x = π/4 and 3π/4. So, option c is correct.
Problem 5 :
The function is concave up in the intervals
-1/2 < x < 0 or x > 1/2
Problem 1 :
If the volume of a cube of side length x is v = x3 . Find the rate of change of the volume with respect to x when x = 5 units.
Problem 2 :
If the mass m(x) (in kilograms) of a thin rod of length x (in meters) is given by, m(x) = √3x then what is the rate of change of mass with respect to the length when it is x = 3 and x = 27 meters
Problem 3 :
A stone is dropped into a pond causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate at 2 cm per second. When the radius is 5 cm find the rate of changing of the total area of the disturbed water?
Problem 4 :
A beacon makes one revolution every 10 seconds. It is located on a ship which is anchored 5 km from a straight shore line. How fast is the beam moving along the shore line when it makes an angle of 45° with the shore?
Problem 5 :
If a ball is thrown into the air with a velocity of 40 ft/s, its height in feet after t seconds is given by
y = -16t2 + 40t
a) Find the average velocity for the time period beginning when t = 2 lasting 0.5 s
b) Find the instantaneous velocity when t = 2.
Problem 6 :
If an arrow is shot upward on the moon with a velocity of 58 m/s, its height in meters after t seconds is given by:
h = 58t - 0.83t2
a) Find the velocity of the arrow at 1 s.
b) Find the velocity of the arrow when t=a.
c) When will the arrow hit the moon?
d) With what velocity will the arrow hit the moon?
Problem 7 :
Temperature readings (in degrees C) where recorded every hour starting at 5:00 am on a day in April in Whitefish, Montana. This table shows some of the readings:
Find the rate of average rate of change for each change in time:
a) 8:00 am to 11:00 am
b) 8:00 am to 10:00 am
c) 8:00 am to 9:00 am
1) 75 units.
2)
When x = 3
m'(3) = (√3/2√3) ==> 1/2 kg/m
When x = 27
m'(27) = (√3/2√27) ==> 1/6 kg/m
3) dA/dt = 20π
4) the beam is moving at the rate of 2π km/sec.
5) Average rate of change = 32
Instantaneous rate of change at x = 2 is -24
6) a) 56.34
b) h'(a) = 58 - 1.66a
c) After 69.8 seconds the
d) -57.868 m/s
7) a) 2
b) 1.95
c) 2.2
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Apr 02, 25 12:40 AM
Apr 02, 25 12:35 AM
Apr 02, 25 12:32 AM