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.

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.

To find the local minima, local maxima, and stationary points, we need to set the derivative equal to zero and solve for x.

The first derivative of f(x) is f'(x) = -5/(x-2)^2.

Setting f'(x) = 0, we have -5/(x-2)^2 = 0.

To solve for x, we can set the numerator equal to zero: -5 = 0.

Since -5 is not equal to zero, there are no values of x that will make the first derivative equal to zero. Therefore, there are no local minima, local maxima, or stationary points for the function f(x) = 2x + 1/(x-2).

Moving on to finding the intervals of upward concavity and downward concavity, we need to analyze the second derivative.

The second derivative of f(x) is f''(x) = 10/(x-2)^2.

We don't want to use the second derivative provided, but we still need to know its value to analyze concavity.

Since the second derivative is positive for all values of x, it means that the function f(x) is always concave up and there are no intervals of downward concavity.

Finally, to find the inflection points, we need to determine where the concavity changes. In this case, since the function is always concave up, there are no inflection points.

In summary:
a) There are no local minima, local maxima, or stationary points.
b) The function is always concave up and has no intervals of downward concavity.
c) There are no inflection points.

I hope this explanation helps you understand how to solve the given exercise, even though you asked to skip the steps for finding the derivatives. Let me know if you have any further questions!