find absolute maximum of function f(x)= e^(-x)/(1+x^2)
(a)1
(b)2
(3)e^-1
(d)e^-1/2
(e)none
2.find minimum value of function f(x)=xlnx
(a)-e
(b)-1
(c)-1/e
(d)e^1/e
(e)none
Note: One of the nicest things to do is to graph the function, if you have that option available. A simple look at the graph will tell you that the answer is e). Failing that possibility:
by the product rule.
We need to find where this is zero.
We only need to numerator to be 0, so:
Divide both sides by (which is never 0):
Now solve for x:
Thus x = -1. (Note: this value of x does NOT make the denominator 0.)
Now, the point (-1, f(-1)) is merely a critical point. It may be a max or a min and it may only be local. It may also be an inflection point. We need to check all these possibilities.
First comment: The function f(x) has no vertical asymptotes.
A reasonable check on whether the critical point is a local or absolute maximum (in the absence of vertical asymptotes) is to take the limit of the function as x goes to positive and negative infinity to see if it "blows up" or "blows down."
Since your function goes to infinity as x approaches to negative infinity, your critical point could only possibly be a local max at best. (As it happens, x = -1 is actually an inflection point, not a local max or min.)
Thus the function has no absolute maximum.
-Dan
I suppose I should finish the problem, even though we have enough information to answer the question.
How do we determine that x = -1 provides an inflection point?
The next step to go to is the second derivative test. The first derivative of f(x) is:
So
If f''(x) > 0 at x = -1 then x = -1 represents a local minimum for the function.
If f''(x) < 0 at x = -1 then x = -1 represents a local maximum for the function.
If f''(x) = 0 at x = -1 then x = -1 represents an inflection point of the function.
Since f''(-1) = 0, x = -1 is an inflection point.
-Dan
I suppose I should finish the problem, even though we have enough information to answer the question.
How do we determine that x = -1 provides an inflection point?
The next step to go to is the second derivative test. The first derivative of f(x) is:
So
If f''(x) > 0 at x = -1 then x = -1 represents a local minimum for the function.
If f''(x) < 0 at x = -1 then x = -1 represents a local maximum for the function.
If f''(x) = 0 at x = -1 then x = -1 represents an inflection point of the function.
Since f''(-1) = 0, x = -1 is an inflection point.
-Dan