What values of x is there an absolute minimum for the function y= -1/3x^2 + 2x?
I found the derivative of the function which is y'=-2/3x + 2
set it to 0=y'=-2/3x + 2
x=3
What do I do now?
what is the second derivative of y?
y"=-2/3, which means that the curve is downward, so that is a maximum...
The curve is a parabola, with a max at (x=3), and has no absolute min.
Thank you!
To determine whether the critical point x = 3 corresponds to an absolute minimum, you need to analyze the concavity of the function.
You can find the second derivative of the function by differentiating the derivative you found:
y'' = d/dx(-2/3x + 2)
= -2/3
Since the second derivative is negative (-2/3 < 0), it means the function is concave down.
By the second derivative test, when a function is concave down and the first derivative changes from positive to negative at a critical point, it indicates that there is an absolute maximum at that point.
In this case, the first derivative changes from positive to negative at x = 3, which means it corresponds to an absolute minimum.
Therefore, x = 3 is the value for which the function y = -1/3x^2 + 2x has an absolute minimum.
Now that you have found the critical value for x, which is x = 3, you need to determine whether this critical value corresponds to an absolute minimum or not.
To determine if it is an absolute minimum, you can use the second derivative test.
1. Compute the second derivative by taking the derivative of the first derivative, y' = -2/3x + 2:
y'' = d/dx (-2/3x + 2) = -2/3
2. Plug the critical value x = 3 into the second derivative:
y'' = -2/3
Since the second derivative is negative for x = 3, this indicates that the function y = -1/3x^2 + 2x has a concave down shape at x = 3.
According to the second derivative test, a concave down shape at a critical point indicates an absolute minimum. Therefore, at x = 3, there is an absolute minimum for the function y = -1/3x^2 + 2x.