Steve
answered
f' = 1 - 1/x
f" = 1/x^2
since f" >0 everywhere, f is concave up everywhere
f'=0 when x=1, so that's a local min.
Step-by-Step Bot
answered
1. Find the first derivative f'(x) of f(x):
f'(x) = 1 - 1/x
2. To determine where the function is increasing or decreasing, set f'(x) = 0 and solve for x:
1 - 1/x = 0
Solve for x: 1 = 1/x
x = 1
Note that x cannot be 0 since ln(0) is undefined.
3. Analyze the sign of f'(x) for x < 1 and x > 1:
For x < 1:
Pick a test point, such as x = 0:
f'(0) = 1 - 1/0 = undefined
Choose another test point, such as x = -1:
f'(-1) = 1 - 1/(-1) = 2
f'(-1) > 0
Therefore, the function is increasing for x < 1.
For x > 1:
Pick a test point, such as x = 2:
f'(2) = 1 - 1/2 = 1/2
f'(2) > 0
Therefore, the function is increasing for x > 1.
4. Determine any relative maximum or minimum:
We can find the sign of the second derivative f''(x) to determine if there are any relative maximum or minimum points. Compute f''(x):
f''(x) = -1/x^2
Since f''(x) is negative for x > 0, there are no relative minimum or maximum points for this function.
In summary:
- The function f(x) = x - ln(x) is increasing for x < 1 and x > 1.
- There are no relative maximum or minimum points for this function.
Explain Bot
answered
Step 1: Find the derivative f'(x) by differentiating each term separately.
f'(x) = d/dx (x - lnx)
= 1 - d/dx (lnx)
Step 2: To differentiate ln(x), recall that d/dx (lnx) = 1/x.
f'(x) = 1 - 1/x
Now that we have the derivative f'(x), we can examine its behavior to determine the intervals where f(x) is increasing or decreasing.
Step 3: Set f'(x) = 0 to find critical points.
1 - 1/x = 0
Step 4: Solve for x by isolating the variable.
1 = 1/x
x = 1
Therefore, x = 1 is a critical point for f(x).
Step 5: Create a sign chart with intervals.
Consider three intervals: x < 1, x = 1, and x > 1.
Interval 1: x < 1
Choose a test point x_1 < 1, for example, x_1 = 0.
Evaluate f'(x_1) = 1 - 1/x_1 = 1 - 1/0 (not defined).
Since f'(x_1) is undefined, we cannot provide information about this interval.
Interval 2: x = 1
Since x = 1 is a critical point, we cannot determine the behavior without further analysis. This point could potentially be a relative maximum or minimum.
Interval 3: x > 1
Choose a test point x_2 > 1, for example, x_2 = 2.
Evaluate f'(x_2) = 1 - 1/x_2 = 1 - 1/2 = 1/2 > 0.
Since f'(x_2) > 0, we can conclude that f(x) is increasing on the interval x > 1, assuming f''(x) exists and is continuous.
Therefore, the function f(x) = x - lnx is increasing on the interval x > 1.
To determine the intervals of decreasing, we can use the same method.
Interval 1: x < 1
Since f'(x_1) is undefined, we cannot provide information about this interval.
Interval 2: x = 1
Since x = 1 is a critical point, we cannot determine the behavior without further analysis. This point could potentially be a relative maximum or minimum.
Interval 3: x > 1
Choose a test point x_2 > 1, for example, x_2 = 2.
Evaluate f'(x_2) = 1 - 1/x_2 = 1 - 1/2 = 1/2 > 0.
Since f'(x_2) > 0, we can conclude that f(x) is decreasing on the interval x > 1.
Therefore, the function f(x) = x - lnx is decreasing on the interval x < 1.
When it comes to determining whether x = 1 represents a relative maximum or minimum point, we need to check the second derivative or perform further analysis.
