Use analytical methods to find the exact global maximum and minimum values of the function f(x)=8x-ln(4x) for x >0. If a global maximum or minimum does not exist, enter the word NONE.
For the global maximum at x=none, But for the Global minimum at x=?
Is the Global minimum 0 or nonw? Any help would be greatly appreciated.
I always encouraged my students to make a rough sketch of the graph.
Pick any 5 or 6 positive x's, then find their y values.
Pick an x close to zero (e.g. x = .00001) and see what y you get. This gives you an indication of the graph does close to zero. (remember x cannot be zero for this function)
In general, the max/min is found by setting the derivative equal to zero, solving for the variable, and then subbing that back into the original equation.
so
f(x) = 8x - ln(4x)
f'(x) = 8 - 1/x
set this equal to zero, you will get x = 1/8
I will let you find the minimum.
y = 8 x - ln (4x)
y' = 8 -4/x
when is that zero?
4/x = 8
x = 1/2
is that a maximum or a minimum?
y" = 0 +4/x^2
for x>0, y" is positive so it is a minimum
so minimum at x = 1/2
what is the value of the function at that minimum?
y = 8(1/2) - ln 2
4 - ln 2 = about 3.3
To find the exact global maximum and minimum values of the function f(x) = 8x - ln(4x), we need to analyze the behavior of the function.
First, let's determine the critical points of the function. The critical points occur where the derivative of the function is equal to zero or undefined.
Taking the derivative of f(x), we have:
f'(x) = 8 - 1/(4x)
Setting f'(x) equal to zero:
8 - 1/(4x) = 0
Solving this equation for x, we have:
1/(4x) = 8
Multiplying both sides by 4x, we get:
1 = 32x
Dividing both sides by 32, we have:
x = 1/32
So, x = 1/32 is a critical point.
Next, let's examine the behavior of the function as x approaches infinity and as x approaches 0.
As x approaches infinity, the natural logarithm term ln(4x) will become very large and positive, which will dominate the function. As a result, the function will tend towards negative infinity.
As x approaches 0, the natural logarithm term ln(4x) will become very large and negative, which will dominate the function. As a result, the function will tend towards positive infinity.
Therefore, since the function tends towards positive and negative infinity as x approaches 0 and infinity respectively, there is no global maximum or minimum.
In conclusion, the exact global minimum does not exist (NONE). However, the Global minimum is not 0. The function tends towards positive infinity as x approaches 0.
To find the global maximum and minimum values of the function f(x) = 8x - ln(4x) for x > 0, we can use calculus.
1. Find the derivative of the function f(x) with respect to x: f'(x) = 8 - 1/x.
2. To find the critical points, we set f'(x) equal to zero and solve for x: 8 - 1/x = 0.
Solving this equation, we get 1/x = 8, which simplifies to x = 1/8.
3. Next, we need to check for the existence of the global maximum and minimum values at the critical point and at the endpoints of the interval x > 0.
a) At the critical point x = 1/8, we can analyze the second derivative to determine if it corresponds to a maximum or minimum.
Taking the second derivative of f(x), we have f''(x) = 1/x^2.
Since f''(1/8) = (1/(1/8))^2 = 64 > 0, this means the second derivative is positive, which indicates a local minimum.
However, we need to check for the global behavior of the function.
b) As x approaches 0 from the right side, f(x) is not defined since the natural logarithm is not defined for x = 0. Therefore, we cannot consider x = 0.
c) As x approaches infinity, f(x) also approaches infinity. Therefore, we do not have a global minimum value.
d) As x approaches infinity, f(x) also approaches infinity. Hence, there is no global maximum value.
In conclusion, the function f(x) = 8x - ln(4x) does not have a global maximum or minimum value for x > 0.