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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.

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Set the derivative equals to 0.

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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

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Smaller x  = 0 ,Larger x  = 5.

b) image

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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

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