How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #f(x) = (x - 1)/x#?

Question

Aurora Alvarado · Accepted Answer

You need its derivative in order to know that.

If we want to know everything about #f#, we need #f'#. Here, #f'(x) = (x-x+1)/x^2 = 1/x^2#. This function is always strictly positive on #RR# without #0# so your function is strictly increasing on #]-oo,0[# and strictly growing on #]0,+oo[#. It does have a minima on #]-oo,0[#, it's #1# (even though it doesn't reach this value) and it has a maxima on #]0,+oo[#, it's also #1#.

Charles Anctil · Answer

undefined To determine where the function $ f(x) = \frac{x - 1}{x} $ is increasing or decreasing and to find the relative maxima and minima:

1. Find the first derivative of the function $ f'(x) $.
2. Set $ f'(x) = 0 $ to find critical points.
3. Determine the sign of $ f'(x) $ in intervals between critical points to identify where the function is increasing or decreasing.
4. Analyze the behavior of the function at critical points to determine relative maxima or minima.

Now, let's proceed with these steps:

1. Find the first derivative of $ f(x) $:
$$ f'(x) = \frac{d}{dx}\left(\frac{x - 1}{x}\right) $$
Using the quotient rule, we get:
$$ f'(x) = \frac{(1)(x) - (x - 1)(1)}{x^2} = \frac{1}{x^2} $$

2. Set $ f'(x) = 0 $ to find critical points:
$$ \frac{1}{x^2} = 0 $$
This equation has no solutions since $ 1/x^2 $ is never zero.

3. Determine the sign of $ f'(x) $ in intervals between critical points:
Since there are no critical points, we consider the intervals of the function.
- When $ x > 0 $, $ f'(x) > 0 $, so the function is increasing.
- When $ x < 0 $, $ f'(x) > 0 $, so the function is increasing.

4. Analyze the behavior of the function at critical points:
Since there are no critical points, there are no relative maxima or minima.

Therefore, the function $ f(x) = \frac{x - 1}{x} $ is increasing for all real numbers $ x $, and it does not have any relative maxima or minima.