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

Answer 1

There are none.

This function has no critical numbers.

#f(x) = (x^2-4)(x^2-2x) = ((x+2)(x-2))/(x(x-2)) = (x+2)/x = 1+2/x#
#f'(x) = -2/x^2# is never #0# and is defined for all #x# in the domain of #f#.

Therefore the function has no critical numbers, so it has no local extrema.

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Answer 2

To find the critical numbers for ( y = \frac{x^2 - 4}{x^2 - 2x} ), you first find the derivative, then set it equal to zero and solve for x. After finding critical numbers, you can use the first or second derivative test to determine whether each critical point corresponds to a maximum, minimum, or neither.

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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.

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.

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.

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.

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