Find T that causes the minimum value of P(t) when P'(t) = 1 - 3e^(0.2sqrt(T))
Then find the minimum value of P(t)
If you look carefully, it seems like it's impossible to exactly express T, but only sqrt(T)...... ? So I'm confused about that.
t = T I apologize for failing to be consistent.
You will see that, after solving for p'(t)'s zero, that sqrt(t) = negative value. That is the true zero of p'(t) but if you square it to get an expression for t, it doesn't work as a zero for p'(t) at all.
Okay, now I understand better ^^
If it's never equal to 0, it doesn't mean it has no minimum value
Think about it : suppose f is defined for . If the derivative is strictly positive, then the function is strictly increasing. This would mean that the minimum value of f is attained for the smallest possible value of x, that is 0.
Because if f is increasing, if a<b, then
So for any , , meaning that is the minimum of f.
Now, you can see, here, that (by taking the second derivative for example)
So the function is strictly decreasing.
So the minimum value of f is attained at the maximum value of T.
If it's infinity, there is no minimum.