You have to remember what the derivatives mean.
To start with, your derivatives are incorrect:
f = (2x+2)/(x-2)
f' = -6/(x-2)^2
f" = 12/(x-2)^3
Fortunately, this small error does not affect their properties in any relevant way.
Now, local max/min are where f'=0 and f"≠0
But, f' is never zero. So, no max/min
f is concave up when f">0. Here, f">0 when x>2
f"<0 when x<2
So, f is concave up on (2,∞) and concave down on (-∞,2)
f has inflection points where f"=0 (changes concavity between up and down).
Here, f" is never zero, so there are no inflection points.
So, how is that possible? Note that f is undefined at x=2, and there is a vertical asymptote there. f changes concavity as it passes the asymptote.
http://www.wolframalpha.com/input/?i=%282x%2B2%29%2F%28x-2%29
Use your skills from pre-calc to sketch the graph a bit, and you can see calculus helps fill in the details.
Use the function f to solve the following:
a) Local minima, local maxima, and stationary points if any. Show work.
b)Intervals of upward concavity and downward concavity if any. Show work.
c) Inflection points if any. Show work.
f(x)=2x+1/x-2
Please, don't do steps for first derivative f'(x)=-5/(x-2)^2
or second derivative
f''(x)=10/(x-2)^2
I need to know how to develop the exercise one you have the derivatives.
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