f(x)= −x2lnx
find the critical points , increasing & decreasing interval and concavity.Please just show me the steps on how to get the increasing and decreasing interval and critical pts . Thanks :)
Just use your derivatives.
f = x^2 lnx
f' = 2x*lnx + x^2 * 1/x = x(2lnx + 1)
f' = 2lnx + 2x*1/x + 1 = 2lnx + 3
Now, f is increasing when f' > 0
f is concave up when f'' > 0
max/min is where f' = 0 and f'' =/= 0
inflection where f'' = 0
Go for it!
To find the critical points, increasing and decreasing intervals, and concavity of the function f(x) = -x^2ln(x), you need to follow these steps:
Step 1: Find the derivative of the function f(x) with respect to x to determine critical points.
Step 2: Determine the intervals where the derivative is positive or negative to find the increasing and decreasing intervals.
Step 3: Find the second derivative of f(x) to determine concavity.
Step 4: Determine the intervals where the second derivative is positive or negative to find the concave up and concave down intervals.
Let's go step by step:
Step 1: Find the derivative of the function f(x):
f(x) = -x^2ln(x)
Using the product rule and the chain rule, the derivative of f(x) can be found as follows:
f'(x) = -2xln(x) - x^2 * (1/x)
Simplifying further, we have:
f'(x) = -2xln(x) - x
Step 2: Determine the intervals where the derivative is positive or negative:
To find the critical points, set f'(x) = 0 and solve for x:
-2xln(x) - x = 0
Taking x as a common factor:
x(-2ln(x) - 1) = 0
This equation is satisfied when either x = 0 or -2ln(x) - 1 = 0.
For x = 0:
f'(0) = -2(0)ln(0) - 0 = 0, but this is an undefined value.
For -2ln(x) - 1 = 0:
-2ln(x) = 1
ln(x) = -1/2
x = e^(-1/2)
So we have one critical point, x = e^(-1/2).
Now, create a sign chart with test intervals to determine the sign of f'(x) in each interval. We need to test three intervals:
Interval 1: (-∞, e^(-1/2))
Pick x = 0 (a number smaller than e^(-1/2)) and substitute it into f'(x):
f'(0) = -2(0)ln(0) - 0 = 0
Since f'(0) = 0, this interval is neither increasing nor decreasing.
Interval 2: (e^(-1/2), 0)
Pick x = 1 (a number between 0 and e^(-1/2)) and substitute it into f'(x):
f'(1) = -2(1)ln(1) - 1 = -1
Since f'(1) = -1, this interval is decreasing.
Interval 3: (e^(-1/2), ∞)
Pick x = 2 (a number greater than e^(-1/2)) and substitute it into f'(x):
f'(2) = -2(2)ln(2) - 2 ≈ -4.386
Since f'(2) ≈ -4.386, this interval is decreasing.
Step 3: Find the second derivative of f(x):
To find the concavity, we need the second derivative. Take the derivative of f'(x):
f''(x) = -2ln(x) - 1 - 2/x
Step 4: Determine the intervals where the second derivative is positive or negative:
To find the concave up and concave down intervals, we set f''(x) equal to zero and solve for x:
-2ln(x) - 1 - 2/x = 0
Multiply through by x to eliminate the fraction:
-2xln(x) - x - 2 = 0
This equation is somewhat complex to solve analytically. It can be solved numerically using methods like Newton's method or graphically using software.
However, we can still analyze the concavity by testing intervals similar to what we did in Step 2:
Interval 1: (-∞, e^(-1/2))
Pick x = 0 (a number smaller than e^(-1/2)) and substitute it into f''(x):
f''(0) = -2ln(0) - 1 - 2/0 = undefined
The second derivative is undefined at x = 0, and we cannot determine concavity for this interval.
Interval 2: (e^(-1/2), ∞)
Pick x = 1 (a number greater than e^(-1/2)) and substitute it into f''(x):
f''(1) = -2ln(1) - 1 - 2/1 = -3
Since f''(1) = -3, this interval is concave down.
Therefore, the critical point and increasing and decreasing interval are as follows:
- Critical point: x = e^(-1/2)
- Increasing interval: None
- Decreasing interval: (-∞, e^(-1/2))
- Concave up interval: None
- Concave down interval: (e^(-1/2), ∞)
Remember to verify this solution with further analysis or graphical methods if required.
Sure! To find the critical points, increasing and decreasing intervals, and concavity of the function f(x) = -x^2ln(x), we'll need to take the following steps:
1. Find the derivative of f(x) with respect to x. Let's call this derivative f'(x).
2. Set f'(x) equal to zero and solve for x. The values of x obtained from this step will be the critical points of the function.
3. Determine the sign of f'(x) in the intervals between the critical points to identify the increasing and decreasing intervals.
4. Find the second derivative of f(x) with respect to x. Let's call this derivative f''(x).
5. Determine the sign of f''(x) in the intervals between the critical points to identify the concavity.
Now let's go through each step in more detail.
Step 1: Find the derivative of f(x):
To find the derivative of f(x), we can use the product rule and the chain rule. The derivative of f(x) = -x^2ln(x) is given by:
f'(x) = -2xln(x) - x^2/x
Simplifying f'(x) further, we get:
f'(x) = -2xln(x) - x
Step 2: Find the critical points:
Set f'(x) = 0 and solve for x:
-2xln(x) - x = 0
Factor out x to get:
x(-2ln(x) - 1) = 0
This equation gives two possibilities:
x = 0 or -2ln(x) - 1 = 0
To solve the second equation, we rearrange it as:
-2ln(x) = 1
Divide both sides by -2:
ln(x) = -1/2
Take the exponential of both sides:
x = e^(-1/2)
So the critical points of f(x) are x = 0 and x = e^(-1/2).
Step 3: Determine the increasing and decreasing intervals:
To find the increasing and decreasing intervals, we need to determine the sign of f'(x) for values of x between the critical points.
Choose a test value, such as x = -1, and substitute it into f'(x):
f'(-1) = -2(-1)ln(-1) - (-1) = 2ln(1) + 1 = 1
Since f'(-1) = 1 > 0, we know that f(x) is increasing to the left of x = 0.
Similarly, for x = 1:
f'(1) = -2(1)ln(1) - (1) = 0
Since f'(1) = 0, we can't determine the sign of f'(x) at that point.
For x = e^(-1/2):
f'(e^(-1/2)) = -2e^(-1/2)ln(e^(-1/2)) - e^(-1/2) = -2e^(-1/2)(-1/2) - e^(-1/2) = -e^(-1/2)
Since -e^(-1/2) < 0, we know that f(x) is decreasing to the right of x = e^(-1/2).
Therefore, the function f(x) is increasing on the interval (-∞, 0) and decreasing on the interval (e^(-1/2), ∞).
Step 4: Find the second derivative of f(x):
To find the second derivative of f(x), we need to differentiate f'(x). Let's call this second derivative f''(x).
f''(x) = -2ln(x) - 1
Step 5: Determine the concavity:
To determine the concavity, we need to determine the sign of f''(x).
For x = 1, substitute the value into f''(x):
f''(1) = -2ln(1) - 1 = -1
Since f''(1) = -1, we know that f(x) is concave down at x = 1.
Similarly, for x = e^(-1/2):
f''(e^(-1/2)) = -2ln(e^(-1/2)) - 1 = -1
Since f''(e^(-1/2)) = -1, we know that f(x) is concave down at x = e^(-1/2).
Therefore, the function f(x) is concave down on the interval (-∞, e^(-1/2)) and at the point x = 1.
To summarize:
- The critical points are x = 0 and x = e^(-1/2).
- The function f(x) is increasing on the interval (-∞, 0) and decreasing on the interval (e^(-1/2), ∞).
- The function f(x) is concave down on the interval (-∞, e^(-1/2)) and at the point x = 1.