Find the extreme values of the function and determine the intervals where the function is increasing/decreasing expressed in interval notation.
x(x-2)^2
So far I've found y'=3x^2-8x+4.
Thanks
extreme where y' = 0
decreasing where y' < 0
y' = (3x-2)(x-2)
so, it should be clear where y'=0.
y' < 0 between the roots.
Easily verify your results using the graph at
http://www.wolframalpha.com/input/?i=x(x-2)%5E2
your derivative is correct. Where is it zero?
0 where x = 2 or x = 2/3
what is second derivative at those points?
y" = 6x - 8
at x = 2, that is POSITIVE so a minimum
at x = 2/3 that is NEATIVE so a maximum
at x ---> +oo that is +oo
at x ---> -oo that is -oo
Now sketch it :)
To find the extreme values and intervals of increase/decrease for the function f(x) = x(x-2)^2, you need to follow these steps:
1. Find the critical points by setting the derivative equal to zero and solving for x.
y' = 3x^2 - 8x + 4 = 0
2. To solve the quadratic equation, you can either factor it or use the quadratic formula. In this case, factoring might be easier. Here's the factoring process:
3x^2 - 8x + 4 = 0
(x - 2)(3x - 2) = 0
Setting each factor equal to zero and solving for x:
x - 2 = 0 --> x = 2
3x - 2 = 0 --> x = 2/3
So, the critical points are x = 2 and x = 2/3.
3. Now, you need to determine whether these critical points correspond to a maximum or minimum point. To do this, you can use the second derivative test, which involves finding the second derivative and evaluating it at each critical point.
The second derivative of f(x) = x(x-2)^2 can be found by differentiating the derivative:
y'' = 6x - 8
Evaluating the second derivative at each critical point:
y''(2) = 6(2) - 8 = 4 > 0 --> the second derivative is positive, indicating a local minimum at x = 2
y''(2/3) = 6(2/3) - 8 = -4 < 0 --> the second derivative is negative, indicating a local maximum at x = 2/3
4. Now that you have identified the nature of the critical points, you can determine the intervals of increase and decrease.
- For x < 2/3, the function is increasing.
- For 2/3 < x < 2, the function is decreasing.
- For x > 2, the function is increasing.
5. Finally, express the intervals of increase/decrease in interval notation:
- Increasing interval: (-β, 2/3) U (2, β)
- Decreasing interval: (2/3, 2)
Therefore, the extreme values of the function are a local maximum at x = 2/3 and a local minimum at x = 2. The function is increasing on the interval (-β, 2/3) U (2, β), and decreasing on the interval (2/3, 2).