SECOND DERIVATIVE TEST FOR LOCAL EXTREMA

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)

Solution

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

Solution

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

Solution

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.

Solution

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

Solution

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?

Solution

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?

Solution

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.

Solution

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?

Solution

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:

rate-of-change-wp-q1

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

Solution

Answer Key

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

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 134)

    Apr 02, 25 12:40 AM

    digitalsatmath143.png
    Digital SAT Math Problems and Solutions (Part - 134)

    Read More

  2. SAT Math Resources (Videos, Concepts, Worksheets and More)

    Apr 02, 25 12:35 AM

    SAT Math Resources (Videos, Concepts, Worksheets and More)

    Read More

  3. Digital SAT Math Problems and Solutions (Part 135)

    Apr 02, 25 12:32 AM

    digitalsatmath147.png
    Digital SAT Math Problems and Solutions (Part 135)

    Read More