extrema points

**bobby77** · October 31st 2006, 11:05 AM

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

**topsquark** · October 31st 2006, 11:26 AM

Originally Posted by bobby77

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

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:

$f(x) = \frac{e^{-x}}{1 + x^2}$

$f'(x) = \frac{-e^{-x}(1 + x^2) - e^{-x}(2x)}{(1+x^2)^2}$ by the product rule.

We need to find where this is zero.
$0 = \frac{-e^{-x}(1 + x^2) - e^{-x}(2x)}{(1+x^2)^2}$

We only need to numerator to be 0, so:
$0 = -e^{-x}(1 + x^2) - e^{-x}(2x)$ Divide both sides by $e^{-x}$ (which is never 0):

$0 = -(1 + x^2) - 2x$

Now solve for x:
$-x^2 - 2x - 1 = 0$

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."

$\lim_{x \to \infty}\frac{e^{-x}}{1+x^2} = 0$
$\lim_{x \to -\infty}\frac{e^{-x}}{1+x^2} \to \infty$

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

**topsquark** · October 31st 2006, 11:39 AM

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:

$f'(x) = \frac{-e^{-x}(1+x^2) - e^{-x}(2x)}{(1+x^2)^2} = - \frac{e^{-x}(x+1)^2}{(1+x^2)^2}$

So
$f''(x) = - \frac{ \left ( -e^{-x}(x+1)^2 + e^{-x}2(x+1) \right )(1+x^2)^2 - e^{-x}(x+1)^2 \cdot 2(1+x^2)2x}{(1+x^2)^4}$

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

**CaptainBlack** · October 31st 2006, 11:41 AM

Originally Posted by topsquark

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:

$f(x) = \frac{e^{-x}}{1 + x^2}$

$f'(x) = \frac{-e^{-x}(1 + x^2) - e^{-x}(2x)}{(1+x^2)^2}$ by the product rule.

We need to find where this is zero.
$0 = \frac{-e^{-x}(1 + x^2) - e^{-x}(2x)}{(1+x^2)^2}$

We only need to numerator to be 0, so:
$0 = -e^{-x}(1 + x^2) - e^{-x}(2x)$ Divide both sides by $e^{-x}$ (which is never 0):

$0 = -(1 + x^2) - 2x$

Now solve for x:
$-x^2 - 2x - 1 = 0$

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."

$\lim_{x \to \infty}\frac{e^{-x}}{1+x^2} = 0$
$\lim_{x \to -\infty}\frac{e^{-x}}{1+x^2} \to \infty$

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

There is no need to look for stationary points for this function, as x-> -infty
the e^(-x) term goes to infinity, and faster than any polynomial in x, so the
whole thing goes to infinity as x->-infty.

RonL

**topsquark** · October 31st 2006, 11:41 AM

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:

$f'(x) = \frac{-e^{-x}(1+x^2) - e^{-x}(2x)}{(1+x^2)^2} = - \frac{e^{-x}(x+1)^2}{(1+x^2)^2}$

So
$f''(x) =$ $- \frac{(-e^{-x}(x+1)^2 + e^{-x}2(x+1))(1+x^2)^2 - e^{-x}(x+1)^2 \cdot 2(1+x^2)2x}{(1+x^2)^4}$

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

**CaptainBlack** · October 31st 2006, 11:49 AM

Originally Posted by bobby77

2.find minimum value of function f(x)=xlnx
(a)-e
(b)-1
(c)-1/e
(d)e^1/e
(e)none

Now this one has its global minimum at a stationary point.

$<br /> \frac{df}{dx}=\ln(x)+1<br />$

so the stationary points are the solutions of:

$<br /> \frac{df}{dx}=\ln(x)+1=0<br />$

which has a single solution at $\ln(x)=-1$ , or $x=e^{-1}$ , where $f(x)=-1/e$

RonL

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