Asked by Alice
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.
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.
Answers
Answered by
Steve
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 hoe calculus helps fill in the details.
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 hoe calculus helps fill in the details.