Hello, please help me :)
A question asked that I find all relative extrema of a function f(x)= x + 1/x. I'm a bit confused with my solution, but I'll show my work first..:
I found f'(x) to get the critical numbers:
f'(x)=1-x^-2
f'(x)=(x^2 -1)/x^2
f'(x)=(x+1)(x-1)/x^2
therefore, critical numbers are x=-1,1 (and x=0 is a vertical asymptote)
so after that, I did a sign diagram using f' to figure out where the function increased/decreased: increase at (-∞,-1) and (1,∞); decrease at (-1,0) and (0,1).
From that I found that my relative max is at a point (-1,-2) and my relative min is at a point (1,2).
Here is where I got a bit confused... If I look at the whole function, my rel. min is larger than my rel. max, but if I look at it, as two different intervals it seems okay, where (-∞,0) would have a rel. max of -2 and (0,∞) would have a rel. min of 2.
Should I not look at the function as a whole? or should I just look at it as two separate open intervals?
Also, my work is okay, correct? Just making sure, I really want to ace this test! :)
Hello! I'll be happy to help you with your question.
First, let's look at your work. You correctly found the derivative of the function f(x) as f'(x) = (x^2 - 1) / x^2. Then you factored f'(x) and found the critical numbers as x = -1, 0, and 1. However, it's important to note that x = 0 is not a critical number because it does not make f'(x) undefined. Instead, it is a point where f'(x) changes from positive to negative or vice versa, indicating a change in concavity.
To determine whether these critical numbers correspond to relative extrema, we can use the first derivative test. If f'(x) changes sign from positive to negative at x = a, then f(a) is a relative maximum. Conversely, if f'(x) changes sign from negative to positive at x = b, then f(b) is a relative minimum.
Now let's analyze the intervals using the sign diagram you constructed for f'(x):
- From (-∞, -1), f'(x) is positive. Therefore, this interval is where f(x) is increasing.
- Between (-1, 0), f'(x) is negative. Thus, f(x) is decreasing in this interval.
- From (0, 1), f'(x) is positive again, implying an increase in f(x).
- For (1, ∞), f'(x) remains positive, indicating continued increase in f(x).
Based on this, we can conclude:
- At x = -1, f(x) changes from increasing to decreasing and represents a relative maximum.
- At x = 0, f(x) changes from decreasing to increasing, indicating a point of inflection rather than a relative extremum.
- At x = 1, f(x) changes from increasing to decreasing and represents a relative minimum.
Therefore, your findings are correct. The relative maximum occurs at (-1, -2), and the relative minimum occurs at (1, 2).
Now, to address your confusion about looking at the function as a whole versus two separate intervals, both viewpoints are valid, but they provide different information. When you consider the entire function, you observe that the relative minimum (2) is greater than the relative maximum (-2). However, if you divide the function into two separate intervals (from -∞ to 0 and from 0 to ∞), you correctly note that the relative maximum in the first interval is -2, while the relative minimum in the second interval is 2. This distinction arises because there is a point of inflection at x = 0, which separates the two intervals and leads to different behavior.
In summary, your work is correct. By considering both the sign diagram and the function as a whole, you correctly identified the relative extrema. Best of luck with your test! If you have any further questions, feel free to ask.