939 said:Homework Statement
Please check my answer. I am weak at math.
Homework Equations
Y = xlnx^2
The Attempt at a Solution
Y' = (lnx^2) + (x)(2x/x^2)
For the max/min, I would just set Y' = 0 by trial and error and then do y", but I don't want to do that until I know y' is correct.
Please help =(
939 said:Homework Statement
Please check my answer. I am weak at math.
Homework Equations
Y = xlnx^2
The Attempt at a Solution
Y' = (lnx^2) + (x)(2x/x^2)
For the max/min, I would just set Y' = 0 by trial and error and then do y", but I don't want to do that until I know y' is correct.
Please help =(
939 said:Thanks for such quick anwers.
So I simplify y' a little bit and have:
y ' = (lnx^2) + (x)(2x/x^2)
y ' = (lnx^2) + (2x^2/x^2)
y ' = (2lnx) + (2x/x)
y = (2lnx) + (2)
But how can I set y ' = 0? 2lnx would have to be - 2, no? It would be invalid, wouldn't it? :-/
Dick said:Not at all. Some numbers have negative logs. Think about that a bit more.
The maximum value of xlnx^2 is achieved when x = e, which results in a maximum value of e*ln(e^2) = 2e.
The minimum value of xlnx^2 is achieved when x = 1, which results in a minimum value of 1*ln(1^2) = 0.
To find the maximum and minimum values of xlnx^2, take the derivative of the function and set it equal to 0. Solve for x to find the critical points, and then plug those values into the second derivative test to determine if they are maximum or minimum values.
For example, if we have the function f(x) = xlnx^2, we can take the derivative: f'(x) = 2lnx + 1. Setting this equal to 0, we get x = e^-1 = 1/e. Plugging this into the second derivative f''(x) = 2/x, we see that it is positive, indicating a minimum value. Therefore, the minimum value of xlnx^2 is f(1/e) = 1/e * ln((1/e)^2) = -2/e.
Yes, since xlnx^2 is a polynomial with a degree higher than 2, it cannot be solved using basic algebraic methods. Calculus is necessary to find the critical points and use the second derivative test to determine the nature of these points.