Let p(x)=x^2(a−x), where a is constant and a greater than 0.
Find the local maxima and minima of p.
(Enter your maxima and minima as comma-separated xvalue,classification pairs. For example, if you found that x=−2 was a local minimum and x=3 was a local maximum, you should enter (-2,min), (3,max). If there were no maximum, you must drop the parentheses and enter -2,min.)
My answer was -2a/3 for the local min.... and that's it, but that's wrong....
yup
To find the local maxima and minima of p(x) = x^2(a - x), we need to take the derivative of p(x) and find its critical points. Let's go through the steps:
1. Find the derivative of p(x) with respect to x using the product rule:
p'(x) = 2x(a - x) + x^2(-1)
= 2ax - 2x^2 - x^2
= 2ax - 3x^2
2. Set p'(x) equal to zero to find critical points:
2ax - 3x^2 = 0
3. Factor out x:
x(2a - 3x) = 0
4. Apply the Zero Product Property:
x = 0 or 2a - 3x = 0
Solving the equation 2a - 3x = 0 for x:
2a = 3x
x = 2a/3
So, we have two critical points: x = 0 and x = 2a/3.
5. To determine if these critical points are local maxima or minima, we need to analyze the second derivative.
Find the second derivative of p(x):
p''(x) = d²/dx² (2ax - 3x^2)
= 2a - 6x
6. Substitute the critical points into p''(x) to determine their classifications.
For x = 0:
p''(0) = 2a - 6(0)
= 2a
If a > 0, then p''(0) > 0, meaning it is a local minimum.
For x = 2a/3:
p''(2a/3) = 2a - 6(2a/3)
= 2a - 4a
= -2a
If a > 0, then p''(2a/3) < 0, meaning it is a local maximum.
Therefore, the local maxima and minima of p(x) = x^2(a - x) are:
(x, classification) = (0, min) and (2a/3, max).
To find the local maxima and minima of the function p(x) = x^2(a−x), where a is a constant greater than 0, we need to take the derivative of p(x) and set it equal to zero. Let's go through the steps:
Step 1: Take the derivative of p(x) with respect to x.
p'(x) = 2x(a−x) + x^2(-1)
= 2ax − 2x^2 − x^2
= 2ax − 3x^2
Step 2: Set p'(x) equal to zero and solve for x.
2ax − 3x^2 = 0
Factor out common terms:
x(2a − 3x) = 0
Solve each factor independently:
x = 0 or 2a − 3x = 0
For x = 0:
p"(0) = 2a(0) − 3(0)^2 = 0
This shows that x = 0 is not a local maximum or minimum.
For 2a − 3x = 0, solving for x:
3x = 2a
x = 2a/3
Step 3: Determine the nature of the critical point using the second derivative test.
To use the second derivative test, we need to find the second derivative of p(x):
p''(x) = 2a − 6x
At x = 2a/3:
p''(2a/3) = 2a − 6(2a/3) = 2a − 4a = -2a
If p''(2a/3) < 0, then it is a local maximum.
If p''(2a/3) > 0, then it is a local minimum.
If p''(2a/3) = 0, then the test is inconclusive.
In this case, p''(2a/3) = -2a, which means that it is inconclusive as it depends on the value of a.
So, the final answer depends on the value of a. If a > 0, then x = 2a/3 is a local minimum. If a = 0 or a < 0, then there are no local maxima or minima.
Therefore, the answer for the local minima and maxima of p(x) = x^2(a−x) is as follows:
- If a > 0: (2a/3, min)
- If a = 0 or a < 0: There are no local maxima or minima.