Critical Numbers





In this page critical numbers we are going to see the definition and procedure to find critical-numbers.

Definition of critical-number:

A Critical-number of a function f is a number c in the domain of f such that either f '(c) = 0 or f ' (c) does not exist.

Definition of stationary points:

Stationary points are critical-numbers c in the domain of f, for which f ' (c) = 0.

Procedure to find critical-number:

  • Find the first derivative
  • set f ' (c) = 0.
  • Solve for c.
  • The value of c are critical-numbers.

Procedure to find Stationary points:

  • Apply those values of c in the original function y = f (x).
  • (x,y) are the stationary points.

Example 1:

Find the critical-numbers and stationary points of the given function y = 4x - x² + 3

Solution:

As per the procedure first let us find the first derivative

y =  f(x) =  4 x - x² + 3

    f ' (x) = 4 (1) - 2 x + 0

    f ' (x) = 4 - 2 x

set f '(x) = 0

   4 - 2 x= 0

   ÷ by 2 ⇒ 2 - x = 0

                 - x = -2

                    x = 2

Therefore the critical number x = 2

Now plug the value of x in the original function y = f (x)

              f (x) =  4 x - x² + 3

              f (2) = 4 (2) - 2² + 3

              f (2) = 8 - 4 + 3

              f (2) = 11 - 4

              f (2) = 7

Therefore the stationary point is (2,7)


Example 2:

Find the critical numbers and stationary points of the given function y = 2 x³ + 3x² - 36 x + 1

Solution:

As per the procedure first let us find the first derivative

y = f (x) = 2 x³ + 3x² - 36 x + 1

    f ' (x) = 2 (3 x²) + 3 (2 x) - 36 (1) + 0

    f ' (x) = 6 x² + 6 x - 36

    f ' (x) = 4 - 2 x

set f '(x) = 0

   6 x² + 6 x - 36 = 0

   ÷ by 6 x² +  x - 6 = 0

        ( x - 2 ) ( x + 3 ) = 0

        x - 2 = 0         x + 3 = 0

             x = 2          x = -3

Therefore the critical number x = 2 and x = -3

Now plug the values of x in the original function y = f (x)

              f (x) =  2 x³ + 3x² - 36 x + 1

substitute x = 2

              f (2) = 2 (2)³ + 3 (2)² - 36 (2) + 1

              f (2) = 2 (8) + 3 (4) - 72 + 1

              f (2) = 16 + 12 -72 + 1

              f (2) = 28 + 1 - 72

              f (2) = 29 - 72

              f (2) = - 43

substitute x = -3

              f (-3) = 2 (-3)³ + 3 (-3)² - 36 (-3) + 1

              f (-3) = 2 (-27) + 3 (9) + 108 + 1

              f (-3) = -54 + 27 + 108 + 1

              f (-3) = -54 + 136

              f (-3) = 82

Therefore stationary points  are  (2,-43) and (-3,42)


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