HOW TO FIND CRITICAL NUMBERS USING DERIVATIVE

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)  =  4x-x²+3

f'(x)  =  4-2 x

set f'(x)  =  0

4-2x  =  0

x  =  2

Therefore the critical number is x = 2.

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

f(x) =  4x-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  =  2x³+3x²-36x+1

Solution :

As per the procedure first let us find the first derivative

y  =  f(x)  =  2x³+3x²-36x+1

f'(x)  =  6x²+6x-36

6x²+6x-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) =  2x³+3x²-36x+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).

Example 3 :

Find the critical points of the function 

f(x)  =  sin²x−cosx

 on the interval (0,2π).

Solution :

f(x)  =  sin²x−cosx

f'(x)  =  2 sinx cos x + sinx

f'(x)  =  sinx(2cos x + sin x)

f'(x)  =  0

sin x (2cos x + 1)  =  0

sin x  =  0 (or)  2cos x + 1  =  0

So, the critical points are π and 2π/3.

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