Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,400 questions

17,804 answers

1,438 comments

40,060 users

Critical points

0 votes

 

 

Question TWO

 

asked Aug 27, 2014 in CALCULUS by zoe Apprentice

3 Answers

0 votes

c) image

Apply derivative with respect of x.

image

image

image

Set the derivative equals to 0.

image

image

image

image

image

image

image

image

Smaller x  = -2, Larger x = 4.

answered Aug 27, 2014 by david Expert
0 votes

To determine critical values of a function

  • find the derivative of the function
  • find the points where the derivative equals to zero.

a)image

image

image

image

image

image

Smaller x  = 0 ,Larger x  = 5.

b) image

image

image

image

image

image

image

image

x  = 0.66666.

 

answered Aug 27, 2014 by david Expert
0 votes

(2).(a).

The cubic function is f(x) = x3 + bx2 + cx and critical point at (1, 1).

f '(x) = 3x2 + 2bx + c.

Substitute the value of x = 1 and f(x) = 1 in the function f(x) = x3 + bx2 + cx.

1 = (1)3 + b(1)2 + c(1)

1 = 1 + b + c

b  = - c.

Substitute the value of x = 1 and f'(x) = 0 in the function f '(x) = 3x2 + 2bx + c.

0 = 3(1)2 + 2b(1) + c

0 = 3 + 2b + c

- 3 = 2b + c  --------> equation 1.

Substitute b = - c in the equation 1.

- 3 = 2(- c) + c

- 3 = - c

c = 3.

b = - c = - 3

The value of b = - 3 and c = 3.

 

(2).(b).

Substitute the values of b = - 3 and c = 3 in the function f(x) = x3 + bx2 + cx.

f(x) = x3 - 3x2 + 3x

f '(x) = 3x2 - 6x + 3

f "(x) = 6x - 6.

Substitute the values of x = 1 in the function f"(x) = 6x - 6.

f "(1) = 6(1) - 6

        = 0

      = Point of inflection.

The third option is correct.

answered Aug 27, 2014 by casacop Expert
edited Aug 27, 2014 by bradely

Related questions

asked Mar 13, 2015 in CALCULUS by anonymous
asked Jul 22, 2014 in CALCULUS by anonymous
asked Oct 11, 2014 in PRECALCULUS by anonymous
asked Oct 25, 2014 in PRECALCULUS by anonymous
...