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How do you find the critical numbers for #f(x)= x^2 - 4# to determine the maximum and minimum?

Answer 1

Ezra Adams

There is one critical point at #(0,-4)# and it is a minimum.

# f(x)=x^2-4 # # f(2)=2^2-4=0# # :. f'(x)=2x #

At a max/min # f'(x)=0=>2x=0 => x=0 #

And, # :. f''(x)=2=> f''(2) > 0 => #minimum graph{x^2-4 [-10, 10, -5, 5]}

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Answer 2

Aurora Alvarado

To find the critical numbers of ( f(x) = x^2 - 4 ), set its derivative ( f'(x) ) to zero and solve for ( x ).

First, find ( f'(x) ): [ f'(x) = 2x ]

Set ( f'(x) ) to zero: [ 2x = 0 ]

Solve for ( x ): [ x = 0 ]

So, the critical number for ( f(x) = x^2 - 4 ) is ( x = 0 ).

To determine if it's a maximum or minimum, you can use the second derivative test or evaluate the sign of the derivative around the critical number:

For the second derivative test: [ f''(x) = 2 ] Since ( f''(x) ) is positive, ( x = 0 ) corresponds to a minimum point.

Thus, ( x = 0 ) is the minimum point of ( f(x) = x^2 - 4 ).

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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.