Find and identify the location of the extreme points of f(x)=8xln4x
To find the extreme points of the function f(x) = 8xln(4x), we need to find the values of x where the derivative of the function equals zero or does not exist. The extreme points can be either a maximum or minimum point.
Step 1: Find the derivative of the function
To find the derivative of f(x), we can use the product rule.
f'(x) = 8x * d/dx(ln(4x)) + ln(4x) * d/dx(8x)
Using the chain rule, the derivative of ln(4x) is (1/(4x)) * 4 = 1/x
f'(x) = 8x * (1/(4x)) + ln(4x) * 8
= 2 + 8ln(4x)
= 2 + 8ln(4) + 8ln(x)
= 2 + 8ln(4) + 8ln(x)
= 2 + 8ln(4) + 8ln(x)
Step 2: Set the derivative equal to zero and solve for x
2 + 8ln(4) + 8ln(x) = 0
Rearranging the equation, we get:
8ln(x) = -2 - 8ln(4)
Dividing by 8, we have:
ln(x) = (-2 - 8ln(4)) / 8
Taking the exponential of both sides, we get:
x = e^((-2 - 8ln(4)) / 8)
Step 3: Evaluate the second derivative to determine the nature of the extreme points
To determine whether the extreme points are maximum or minimum points, we need to evaluate the second derivative.
f''(x) = (d/dx)(2 + 8ln(4) + 8ln(x))
= 8/x
Step 4: Analyze the second derivative
Since f''(x) = 8/x, and for any value of x, 8/x is positive, we can conclude that f''(x) > 0 for all x. This means that the function has a concave-upward shape, indicating that we have a minimum point.
Step 5: Find the final extreme point
Substitute the value of x from Step 2 into the original function f(x) = 8xln(4x) to get the y-coordinate of the extreme point.
x = e^((-2 - 8ln(4)) / 8)
y = 8xln(4x)
Finally, you can evaluate the y-coordinate using a calculator or numerical methods to find the exact value of the extreme point.
Note: The process can be simplified if you can use a graphing calculator or software that can automatically find the critical points.