Find the points at which y = f(x) = 5x−ln(5x) has a global maximum, a global minimum, and a local, non-global maximum on the interval .1 ≤ x ≤ 2. Round your answers to two decimal places.
Global Minimum:
(x,y) = (,)
Global Maximum:
(x,y) = (,)
Local, Non-Global Maximum:
(x,y) = (,)
f'(x) = 5 - 1/x
for local max/min, set 5-1/x = 0
x =1/5
find f(1/5) for the local max/min
now evaluate
f(.1) , f(2) to find the max/min at the end points.
To find the points where the function y = f(x) = 5x − ln(5x) has a global minimum, global maximum, and local, non-global maximum on the interval .1 ≤ x ≤ 2, we will follow these steps:
1. Find the critical points of the function by taking the derivative and setting it equal to zero.
2. Determine the endpoints of the interval.
3. Evaluate the function at the critical points and endpoints to find the global minimum, global maximum, and local maximum.
Step 1: Find the critical points
To find the critical points, we need to take the derivative of the function:
f'(x) = 5 - 1/(5x)
Setting this equal to zero, we get:
5 - 1/(5x) = 0
1/(5x) = 5
1 = 25x
x = 1/25
Step 2: Determine the endpoints of the interval
The given interval is .1 ≤ x ≤ 2. The endpoints of this interval are 0.1 and 2.
Step 3: Evaluate the function at the critical points and endpoints
To find the y-values, we substitute the x-values into the function:
f(0.1) = 5(0.1) − ln(5(0.1)) ≈ -0.267
f(2) = 5(2) − ln(5(2)) ≈ 9.614
Evaluate the function at the critical point:
f(1/25) = 5(1/25) − ln(5(1/25)) ≈ -0.647
The global minimum occurs at (0.1, -0.267).
The global maximum occurs at (2, 9.614).
The local, non-global maximum occurs at (1/25, -0.647).
Therefore, the answers rounded to two decimal places are:
Global Minimum: (0.1, -0.27)
Global Maximum: (2, 9.61)
Local, Non-Global Maximum: (0.04, -0.65)
To find the points at which y = f(x) = 5x - ln(5x) has a global maximum, a global minimum, and a local, non-global maximum on the interval .1 ≤ x ≤ 2, we need to follow these steps:
Step 1: Calculate the derivative of f(x) with respect to x to find the critical points.
The derivative of f(x) = 5x - ln(5x) can be calculated using the power rule and the chain rule:
f'(x) = 5 - 1/(5x) = 5 - 1/5x
Step 2: Set f'(x) equal to zero and solve for x to find the critical points.
Setting f'(x) = 5 - 1/5x = 0, we can solve for x:
5 - 1/5x = 0
1/5x = 5
x = 1
Step 3: Determine the nature of the critical points.
To determine the nature of the critical points, we need to analyze the second derivative of f(x) with respect to x.
The second derivative of f(x) = 5x - ln(5x) can be calculated as follows:
f''(x) = (d/dx)(5 - 1/5x) = 1/5x^2
Step 4: Analyze the second derivative at the critical points.
Substituting x = 1 into f''(x) = 1/5x^2, we get:
f''(1) = 1/5(1)^2 = 1/5 > 0
Since f''(1) > 0, the critical point x = 1 corresponds to a local minimum.
Step 5: Determine the values of f(x) at the critical points and endpoints of the interval.
At x = .1, f(x) = 5(.1) - ln(5(.1)) ≈ 0.923
At x = 2, f(x) = 5(2) - ln(5(2)) ≈ 7.39
At x = 1 (critical point), f(x) = 5(1) - ln(5(1)) = 5 - ln(5) ≈ 2.61
Step 6: Compare the values to find the global maximum, global minimum, and local, non-global maximum.
From the values calculated above:
- The smallest value occurs at x = .1, with f(x) ≈ 0.923. This corresponds to the global minimum.
- The largest value occurs at x = 2, with f(x) ≈ 7.39. This corresponds to the global maximum.
- At x = 1 (the critical point), f(x) ≈ 2.61. This corresponds to the local, non-global maximum.
Rounding the answers to two decimal places, we have:
Global Minimum: (x,y) ≈ (.10, 0.92)
Global Maximum: (x,y) ≈ (2.00, 7.39)
Local, Non-Global Maximum: (x,y) ≈ (1.00, 2.61)