FINDING LOCAL EXTREMA WITH SECOND DERIVATIVE TEST

The Second Derivative Test relates the concepts of critical points, extreme values, and concavity to give a very useful tool for determining whether a critical point on the graph of a function is a relative minimum or maximum. 

Suppose that c is a critical point at which

f'(c) = 0

that f ′(x) exists in a neighborhood of c, and that f ′(c) exists.

Then f has a

  • relative maximum value at c if f''(c) < 0 and
  •  relative minimum value at c if f''(c) > 0 
  • If f ''(c) = 0 , the test is not informative

Finding Concavity from Second Derivative

(i)  If f''(x) > 0 on an open interval I, then f(x) is concave up on I.

(i)  If f''(x) < 0 on an open interval I, then f(x) is concave down on I.

Find the local extrema for the following functions using second derivative test :

Problem 1 :

f(x) = −3x5+ 5x3

Solution :

f(x) = −3x5+ 5x3

f'(x)  =  -15x4 + 15x2

f'(x)  =  0

-15x2(x2 - 1)  =  0

x  =  0 and x2 - 1  =  0

x  =  0, 1 and -1

f''(x)  =  -20x3+30x

f''(0)  =  -20(0)3+30(0)  ==>  0

f''(1)  =  -20(1)3+30(1)  ==>  10 > 0 (local minimum)

f''(-1)  =  -20(-1)3+30(-1)  ==>  -10 < 0 (local maximum)

So, local maximum at x  =  -1.

Maximum value :

f(-1)  =   −3(-1)5+ 5(-1)3

f(-1)  =   3 - 5  ==>  -2

So, local maximum point is (-1, -2) and local maximum is -2.

So, local minimum at x  =  1.

Minimum value :

f(1)  =   −3(1)5+ 5(1)3

f(1)  =   -3 + 5  ==>  2

So, local maximum point is (1, 2) and local maximum is 2.

Problem 2 :

f(x)  =  x logx

Solution :

f(x)  =  x logx

f'(x)  =  x(1/x) + log x (1)

f'(x)  =  1 + log x

f'(x)  =  0

1 + log x  =  0

log x  =  -1

x  =  e-1

x  =  1/e

f''(x)  =  0 + 1/x

f''(x)  =  1/x

f''(1/e)  =  e > 0 (local minimum)

Local minimum at x  =  1/e

f(1/e)  =  (1/e) log (1/e)

f(1/e)  =  (1/e) log (e-1)

f(1/e)  =  (1/e) (-1)

f(1/e)  =  -1/e

So, local minimum is -1/e

Problem 3 :

f(x)  =  x2e-2x 

Solution :

f(x)  =  x2e-2x 

f'(x)  =  x2(-2e-2x) + e-2x(2x)

e-2x (-2x2+2x)  =  0

e-2x (x-x2)  =  0

x(1-x)  =  0

x  =  0 and x  =  1

f''(x)  =  x(-1) + (1-x)(1)

f''(x)  =  -x + 1-x

f''(x)  =  1-2x

f''(0)  =  1 > 0 (Local minimum)

f''(1)  =  -1 (Local maximum)

Local minimum :

At x  =  0

f(x)  =  x2e-2x 

f(0)  =  0

Local maximum :

At x  =  1

f(1)  =  12e-2(1) 

f(1)  =  1/e2

Local maximum  =  1/e2

Local minimum  =  0

Problem 4 :

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 :

f(x)  =  4x3+3x2-6x+1

f'(x)  =  12x2+6x-6

f'(x)  =  0

12x2+6x-6  =  0

6(2x2+x-1)  =  0

(2x-1)(x+1)  =  0

x  =  1/2 and x  =  -1

The function is increasing on (-∞, -1) and (1/2, ∞), it is decreasing on (-1, 1/2).

f'(x)  =  (2x2+x-1) 

f''(x)  =  4x + 1

f''(x)  =  0

x  =  -1/4

 (-∞, -1/4)  and (-1/4, ∞) 

From  (-∞, -1/4), we take -2

f''(-2)  =  -7 < 0 (Local maximum)

Concave downward on (-∞, -1/4).

From (-1/4, ∞) , we take 2

f''(2)  =  9 < 0 (Local minimum)

Concave upward on (-1/4, ).

At -2 it changes from negative to positive, then it turns into maximum to minimum.

Local maximum :

x  =  -1

f(-1)  =  4(-1)3+3(-1)2-6(-1)+1

=  -4+3+6+1

=  6

Local minimum :

x  =  1/2

f(1/2)  =  4(1/2)3+3(1/2)2-6(1/2)+1

=  (1/2)+(3/4)-3+1

=  (2+3-8)/4

=  -3/4

Point of inflection :

f(1/4)  =  4(-1/4)3+3(-1/4)2-6(-1/4)+1

=  -1/16 + 3/16 + 6/4 + 1

=  (-1+3+24+16)/16

=  42/16

=  21/8

Point of inflection is (-1/4, 21/8)

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