What are the critical points of #f(x) =(x^24x+5)/(x3)#?
The critical numbers are
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To find the critical points of ( f(x) = \frac{{x^2  4x + 5}}{{x  3}} ), you need to first find its derivative, set it equal to zero, and solve for ( x ). Then, check for any potential vertical asymptotes.

Find the derivative of ( f(x) ) using the quotient rule: [ f'(x) = \frac{{(x  3)(2x  4)  (x^2  4x + 5)(1)}}{{(x  3)^2}} ]

Simplify ( f'(x) ): [ f'(x) = \frac{{2x^2  10x + 12  x^2 + 4x  5}}{{(x  3)^2}} ] [ f'(x) = \frac{{x^2  6x + 7}}{{(x  3)^2}} ]

Set ( f'(x) ) equal to zero and solve for ( x ): [ x^2  6x + 7 = 0 ] [ (x  1)(x  7) = 0 ] [ x = 1 \quad \text{or} \quad x = 7 ]

Check for vertical asymptotes by identifying any values of ( x ) that make the denominator of ( f(x) ) equal to zero: [ x  3 = 0 ] [ x = 3 ]
So, the critical points of ( f(x) = \frac{{x^2  4x + 5}}{{x  3}} ) are ( x = 1 ) and ( x = 7 ), and there is a potential vertical asymptote at ( x = 3 ).
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When evaluating a onesided 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 onesided 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 onesided 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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