Find all the critical points of the function
TexFormula1" title=" f(x)= (x+1)/(x-3) " alt=" f(x)= (x+1)/(x-3) " align="absmiddle" class="latex-formula"> and determine the intervals in which the function is increasing and in which it is decreasing?
1 answer:
The function
... y = 1/x
has derivative
... y' = -1/x²
which has no zeros. It is undefined at x=0, the only critical point. The derivative is negative for all values of x, so the function is decreasing everywhere in its domain.
Your function
... y = (x+1)/(x-3)
can be written as
... y = 1 +4/(x-3)
which is a version of y = 1/x that has been vertically scaled by a factor of 4, then shifted 1 unit up and 3 units to the right. Shifting the function to the right means x=3 is excluded from the domain (and the interval on which the function is decreasing).
The critical point is x=3.
The function is decreasing on (-∞, 3) ∪ (3, ∞), increasing nowhere.
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Answer:
None
Step-by-step explanation:
<em><u>Parallel </u></em>means exactly the same just lower or higher(same slope)
<u><em>Perpendicular </em></u>means they come together at a right angle, it is evident these do not
<em><u>Coincident </u></em>means directly on top of one another which they are not
Since it is none of these than the answer is obviously none of these ;)
I hope this helps u pls give a brainliest and a thx :)
<h2>SOLVING</h2>
How do the graphs f(x) and f^-1(x) relate?
These two functions are inverses of each other.
That's the way these two are related to each other.
f^-1(x) is the inverse function of f(x), and f(x) is the inverse function of f^-1(x).