Find the critical points of f(x)=((x+1))/(x-3).

Quinlan7g 2022-09-18 Answered
Finding the critical points of a function and the interval where it increases and decreases.
I am having trouble finding the critical points of
f ( x ) = ( x + 1 ) / x 3
I found the derivative to be f ( x ) = 4 / ( x 3 ) 2 .
My next step was to equate my derivative to zero, but that does not seem to work as my x cancels out. Usually I would take the x-value(worked out by equating the derivative with zero) and substitute it into the original equation to get a y-value. This would then be the critical points. Is there anyone who could maybe help me out (maybe with an example or so) as I also have to find the intervals where the function is increasing and decreasing?
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Answers (1)

nutnhonyl8
Answered 2022-09-19 Author has 8 answers
Step 1
Judging by the derivative you calculated, it appears the function is supposed to be f ( x ) = x + 1 x 3 .
Since the implicit domain of a rational function is the set of all real numbers except those that make the denominator equal to zero, the implicit domain of f is Dom f = { x R x 3 } = ( , 3 ) ( 3 , ).
The derivative of f is f ( x ) = 4 ( x 3 ) 2 < 0 for every x in its domain, which tells us that the function is decreasing on the intervals ( , 3 ) and ( 3 , )
Note that f ( x ) = x + 1 x 3 = 1 + 4 x 3 so lim x f ( x ) = lim x ( 1 + 4 x 3 ) = 1 lim x 3 + f ( x ) = lim x 3 + ( 1 + 4 x 3 ) = lim x 3 + f ( x ) = lim x 3 + ( 1 + 4 x 3 ) = lim x f ( x ) = lim x ( 1 + 4 x 3 ) = 1
Step 2
Thus, when x ( , 3 ), f ( x ) ( , 1 ), and when x ( 3 , ), f ( x ) ( 1 , ). Since the function assumes larger values in the interval ( 3 , ) than it does in the interval ( , 3 ) than it does in the interval ( , 3 ), it does not decrease over the union of the two intervals in which it is decreasing.
If you define a critical point of a function f to be a point where f ( x ) = 0, then the function f : ( , 3 ) ( 3 , ) R defined by
f ( x ) = x + 1 x 3 does not have a critical point since f ( x ) < 0 for every x in its domain.
If you use the alternative definition that a critical point of a function f is a point in its domain where f ( x ) = 0 or f′(x) does not exist, then the function f : ( , 3 ) ( 3 , ) R defined by f ( x ) = x + 1 x 3 still does not have a critical point since the derivative is defined at every point of its domain.

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