Mth 173
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Query problem 4.2.37 (3d edition 4.2.31) U = b( a^2/x^2 - a/x)
I said in my homework: This point (2a, -b/4) is a minimum, which can only be determined by graphing it. The first and second derivative tests don’t help because we don’t know whether a is < 1 or not.
If a is < 1 and small enough, then the second derivative will be negative, not positive, making the point a maximum. The same problem holds for the first-derivative test. So, we just have to graph the function using the asymptotes and x intercepts in order to see that the point is a minimum.
I'm not sure about this. What am I missing?
The derivative of your function is -2 a^2 / x^3 + a / x^2. Setting this equal to zero, and multiplying through by x^3 we get the equation
-2 a^2 x + a x^2 = 0, which we solve to get
x = 2a or x = 0.
x = 0 isn't in the domain of the original equation, so we reject this.
Thus (2a, -b/4) is a critical point.
We can use a first- or second-derivative test to determine whether this is a max, a min or perhaps an inflection point.
The second derivative of our function is
6 a^2 / x^4 - 2 a / x^3.
When x = 2 a the second derivative is zero, a^2 / 8, which is positive.
So as long as b is positive, The graph at this point is concave up and the critical point gives us a minimum.
The second derivative does become 0 at x = 3a, and the function changes concavity at that point. However that point is not a critical point.