find where f(x)<0 and f(x)>0
f(x)=xlnx-2x
f(x) = 0 when lnx = 2
x = e^2 = 7.389 when f(x) = 0
f(x)> 0 for x > e^2
f(x) <0 for x < e^2
x*ln(x)-2x<0
x*ln(x)<2x Divided with x
ln(x)<2
e^2=7.389 x<7.389
x*ln(x)-2x>0
x*ln(x)>2x Divided with x
ln(x)>2
e^2=7.389 x>7.389
Why post the same answer over 3 hours later?
To find where f(x) is less than zero (f(x) < 0) and greater than zero (f(x) > 0), we need to examine the sign of the function for different values of x.
Let's start by finding where f(x) = 0, as it will help us identify where the function changes sign.
Step 1: Find where f(x) = 0.
Setting f(x) = 0, we have:
xlnx - 2x = 0
To solve this equation, we can factor out an x:
x(lnx - 2) = 0
This equation is satisfied when either x = 0 or lnx - 2 = 0.
For x = 0, f(0) is not defined, so we exclude it from further analysis.
Solving lnx - 2 = 0:
lnx = 2
x = e^2
So, f(x) = 0 at x = e^2.
Step 2: Determine the intervals where f(x) is positive or negative.
Now, we need to examine the sign of f(x) in regions defined by the intervals.
To do this, we select test points in each interval and evaluate the function.
Let's select values in the following intervals:
Interval 1: x < 0
Interval 2: 0 < x < e^2
Interval 3: x > e^2
Select a point x1 in Interval 1:
Let x1 = -1 (arbitrary value less than 0)
f(x1) = -1 * ln(-1) - 2(-1)
Since ln(-1) is not defined, we cannot determine the sign of f(x1) in this interval.
Select a point x2 in Interval 2:
Let x2 = 1 (arbitrary value between 0 and e^2)
f(x2) = 1 * ln(1) - 2(1)
f(x2) = 0 - 2 < 0
So, f(x) is negative (f(x) < 0) in Interval 2.
Select a point x3 in Interval 3:
Let x3 = 3 (arbitrary value greater than e^2)
f(x3) = 3 * ln(3) - 2(3)
Since both ln(3) and 3 are positive, f(x3) > 0.
Therefore, we conclude that f(x) is negative (f(x) < 0) in the interval (0, e^2) and positive (f(x) > 0) in the interval (e^2, ∞).
In summary:
f(x) < 0: (0, e^2)
f(x) > 0: (e^2, ∞)