f(x) = 1/x + ln(x), 0.5<=x<=4
The local maximum of this function is given by (1/2, 2-ln(2))
How do we get this answer?
It is easiest enough to graph and see that is the endpoint, and the graph is coming from a maximum.
To find the local maximum of the function f(x) = 1/x + ln(x), within the given interval [0.5, 4], we need to follow these steps:
Step 1: Find the critical points
A critical point occurs when the derivative of the function is equal to zero or undefined. We need to find the values of x that make f'(x) = 0 or f'(x) undefined.
First, let's find the derivative of f(x):
f'(x) = d/dx(1/x + ln(x))
To find the derivative of 1/x, we can use the power rule:
d/dx(1/x) = -1/x^2
To find the derivative of ln(x), we can use the chain rule:
d/dx(ln(x)) = 1/x
Now, let's combine the derivatives:
f'(x) = -1/x^2 + 1/x
Next, set f'(x) equal to zero:
-1/x^2 + 1/x = 0
Multiply both sides by x^2 to eliminate the denominators:
-x + x^2 = 0
Rearrange the equation to get:
x^2 - x = 0
Factor x out:
x(x - 1) = 0
This equation gives us two critical points: x = 0 and x = 1. However, the interval given is [0.5, 4], so we can eliminate x = 0 since it is outside the interval.
Step 2: Evaluate the function at the critical points and endpoints
Now, we need to evaluate the function f(x) at the critical points (x = 1) and the endpoints of the interval (x = 0.5 and x = 4).
f(0.5) = 1/0.5 + ln(0.5) = 2 - ln(2) ≈ 0.6137
f(1) = 1/1 + ln(1) = 1 + 0 = 1
f(4) = 1/4 + ln(4) = 0.25 + ln(4) ≈ 1.3863
Step 3: Identify the local maximum
Comparing the values of f(x) at x = 0.5, x = 1, and x = 4, we can see that f(0.5) is the highest value, which is approximately 0.6137. Therefore, the local maximum of f(x) within the given interval [0.5, 4] is at (1/2, 2 - ln(2)).
Hence, the local maximum of the function f(x) = 1/x + ln(x), within the interval [0.5, 4], is given by (1/2, 2 - ln(2)).