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How do you find the relative extrema for #f(x) = (x^2 - 3x - 4)/(x-2)#?

Answer 1

Charles Arocha

That function has no relative extrema.

#f(x) = (x^2 - 3x - 4)/(x-2)#

Domain of #f# is #(-oo,2)uu(2,oo)#

#f'(x) = (x^2 - 4x +10)/(x-2)^2#

#f'(x)# exists for all #x# in the domain of #f#

#f'(x) = 0# where #x^2-4x+10 = 0#. But this quadratic has discriminant #16-40# which is negative. Therefore, there are no real solutions to the equation, and consequently there are no critical numbers for #f#.

Since relative extrema occur at critical numbers, there are no relative extrema for #f#.

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Answer 2

Elijah Abram

To find the relative extrema of the function f(x) = (x^2 - 3x - 4)/(x-2), you first find the derivative f'(x) using the quotient rule. Then, you set f'(x) equal to zero to find critical points. Finally, you test these critical points to determine whether they correspond to relative extrema.

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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.