Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

How do you find the critical numbers of #f(x)=x^2-6x#?

Answer 1

Critical point exist at #x=3#

#f(x) = x^2 - 6x#

Critical points can be found at x = c where #f'(c) = 0# critical points are also obtained at #x= c# if #f(c)# is defined and #f'(c)# is not defined.

So first we find derivative #f'(x) #

#f'(x) = 2x - 6#

We find x by solving #f'(x)=0#

#2x - 6 = 0# #=> 2x = 6 # #=> (2x)/2 = 6/2#

#=> x= 3# Critical point exists at this value of x.

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Answer 2

Isaiah Allen

To find the critical numbers of ( f(x) = x^2 - 6x ), we first find its derivative, ( f'(x) ), then set it equal to zero and solve for ( x ).

So, the derivative of ( f(x) ) is ( f'(x) = 2x - 6 ).

Setting ( f'(x) ) equal to zero, we get:

( 2x - 6 = 0 )

Solving for ( x ), we have:

( 2x = 6 ) ( x = 3 )

Therefore, ( x = 3 ) is the critical number of ( f(x) = x^2 - 6x ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.