Does the given function have a minimum and maximum?

[mathboy20]

Hi

I have been given the following assignment which has caused me some trouble:

The function $$f(x) = x^b \cdot e^{-x}$$ where $$b \in \mathbb{R}_{+}$$

Determine if f has a minimum and a maximum, and find them.

I know that the first step is determine f'(x) which is

$$f'(x) = (\frac{b}{x} - ln(e)) \cdot e^{-x} \cdot x^b$$

Any hints what I do next ?

Best Regards

Mathboy20

 

[TD]

I know that the first step is determine f'(x) which is

$$f'(x) = (\frac{b}{x} - ln(e)) \cdot e^{-x} \cdot x^b$$

Any hints what I do next ?

If there is a minimum or a maximum, the derivative has to be zero there. Watch out though, a zero derivative doesn't necessarily imply an extremem (min or max) but it is a necessary condition. So find the values of x for which $f'(x)=0$.

 

[mathboy20]

Do I then choose an abitrary b-value and then solve f'(x) = 0 ?

Like let's say b = 1

then $$f'(x) = -(ln(e) \cdot x -1) \cdot e^{-x}= 0$$

Then $$x = \frac{1}{ln(e)}$$

If I then choose an abitratry b which lies in the inteval $$[1, \infty[$$

I get that $$x = \frac{b}{ln(e)}$$

is that x then the maximum value for the given function?

Best Regards
Mathboy20

If there is a minimum or a maximum, the derivative has to be zero there. Watch out though, a zero derivative doesn't necessarily imply an extremem (min or max) but it is a necessary condition. So find the values of x for which $f'(x)=0$.

 

[TD]

Rather then doing it for a particular value of b, do it in general (keep b as a parameter).
You indeed get $x = \frac{b}{\ln(e)}$ but don't you think you could simplify $\ln(e)$ a bit?

 

[mathboy20]

Rather then doing it for a particular value of b, do it in general (keep b as a parameter).
You indeed get $x = \frac{b}{\ln(e)}$ but don't you think you could simplify $\ln(e)$ a bit?

Yeah sure then x = b

Can I then conclude that the parameter 'b' is the maximum value for the function ?

Best Regards

Mathboy20

 

[VietDao29]

Yeah sure then x = b

Can I then conclude that the parameter 'b' is the maximum value for the function ?

Best Regards

Mathboy20

Be careful, when b is not 1, then there is one more value that makes f'(x) = 0. That value is x = 0. Do you know why?
To check if it's the maximum or minimum value, one can try to take the second derivative of that function.
If f''(x) > 0, and f'(x) = 0, then it's a minimum value.
If f''(x) < 0, and f'(x) = 0, then it's a maximum value.
Do you know this?

 

[TD]

Yeah sure then x = b

Can I then conclude that the parameter 'b' is the maximum value for the function ?

Although it's correct (this will be a maximum), you cannot conclude this from the zero derivative only. In order for that point to be a max (c.q. min), the derivative has to change sign arround that point (from + to - for a max and vice versa for a min).

Alternatively, you could check the sign of the second derivative in that point. If that's negative, you've got a maximum while you'll have a minimum when that's positive.

 

[mathboy20]

Hello again,

Lets recap what I know.

I'm given the function $$f(x) = x^{a} \cdot e^{-x}$$

where $$a \in \mathbb{R}_{+}$$ which means that $$a > 0$$

Futher I'm told that $$x \in [0, \infty[$$

I required first to prove that there exists a minimum and a maximum value for f, and to determain those values.

What would be the first logical step here?

Sincerely Yours
Mathboy20

 

[VietDao29]

First, you should notice that:
f(x) > 0 for all x > 0, and f(x) = 0 for x = 0, right? So what can you say about f(0)? (a maximum value or minimum value?)
And the second hint is to follow TD's suggestion and prove that x = b is a maximum.
Can you go from here? :)

 

[mathboy20]

Dear TD,

If I understand Your explanation correctly then I get the following

first my f'(x) was wrong the right one is

f'(x) = (b/x -1) * x^b * e^-x =0

Then according to the definition then there exist a maximum if f''(x) < 0 and f'(x) = 0

first condition:

by choosing x = b

then f'(b) = 0

Second condition:

f''(b) = -b ^ (b-1) * e^-b < 0 because b>0

Then the maximum for f is x = b...

There exist a minimum if f''(x) > 0 and f'(x) = 0

The first point in the definition interval is x = 0, but this is unuserable since b/0 is not allowed.

Then f doesn't have a mimimum?

Sincerley Yours
Mathboy20

 

[VietDao29]

f'(x) = bx^{b - 1}e^-x - x^be^-x
-------------------
f''(x) = b(b - 1)x^{b - 2}e^-x - bx^{b - 1}e^-x - bx^{b - 1}e^-x + x^be^-x
= e^-x(b(b - 1)x^{b - 2} - 2bx^{b - 1} + bx^b)
f''(b) = be^-b(b(b - 1)b^{b - 2} - 2bb^{b - 1} + b^b) = e^-b(-b^b - b^{b - 1} + b^b) = -b^{b - 1}e^-b < 0.
Yes, so your work is correct.
-------------------------
However, you should also check the value of f(x) at the end point, too to see if it's a maximum or minimum there. As I told you before:
f(0) = 0
And f(x) > 0 for all x > 0
So f(0) is a minimum value for f(x), since the domain of the function is $$[0 ; \infty [$$
By the way, f'(x) = bx^{b - 1}e^-x - x^be^-x = x^{b - 1}e^-x(b - x)
So as long as b is not 1, f'(0⁺) = 0.