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)).