# Critical points

Question TWO

asked Aug 27, 2014 in CALCULUS

c)

Apply derivative with respect of x.

Set the derivative equals to 0.

Smaller x  = -2, Larger x = 4.

To determine critical values of a function

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

a)

Smaller x  = 0 ,Larger x  = 5.

b)

x  = 0.66666.

(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.