Function x - Finding the intervals, ranges, and points of inflection

In summary: Yeah, that sounds right. :) In summary, the function f has an increasing interval from x =-1 to x = 1, and a range of -1 to 1.
  • #1
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Homework Statement



Let f be the function defined by f(x) = xe[tex]^{1-x}[/tex] for all real numbers x.

a. Find each interval on which f is increasing.

b. Find the range of F.

c. Find the x-coordinate of each point of inflection of the graph of f


Homework Equations



I don't think there are any.

The Attempt at a Solution



It's been a while since I did calculus, so I might be a bit rusty on this. I'll take this 1 letter at a time.

For part a, if f is incresaing, then its first derivative must be positive. Therefore, I took the derivative of f(x), using the product rule.

f(x) = xe[tex]^{1-x}[/tex]
f'(x) = x(1-x)(-1) + e[tex]^{1-x}[/tex]
f'(x) = -x[tex]^{2}[/tex]+x-1+e[tex]^{1-x}[/tex]

And ... this is where I get stuck. I'm not sure how to proceed from here for part a (I'll attempt the other 2 at a later date). :cry:
 
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  • #2
The product rule is:
(f*g)' = f'*g + f*g' where f and g are functions.
Try finding f'(x) again.
 
  • #3
My notes say it's f*g' + g*f'. Or uv' + vu'.

I've used this rule for other prod. rule problems and it seems to work all right.
 
  • #4
It doesn't matter what the order is; addition is commutative.

Your error was when differentiating e1 - x.
 
  • #5
I thought e^(1-x) was -(1-x) because I bring down the (1-x) from the expnoent, then use the chain rule to get (-1) and just multiply the two together. Is it just (1-x)?
 
  • #6
You're missing a part when you differentiate e1-x. What is the derivative of ef(x) using the chain rule?
 
  • #7
Wait hang on ... is it ...

f'(x) = -e[tex]^{1-x}[/tex]
 
  • #8
Yes!
Now what is f'(x), all of it?
 
  • #9
Wow, I nearly ate my shorts. :biggrin:

Anyway, here's the new f'(x).

f ' (x) = e[tex]^{1-x}[/tex](1-x)

I'm still not quite sure what to do next, can you sort of "poke" me in the general direction without actually giving away the answer since I'd like to learn this stuff.
 
  • #10
Now you want to find where f(x) changes from increasing to decreasing, or vice versa. What will the derivative be right where that change is?
 
  • #11
Just a quick question, but is finding what you said possible without a calculator, becuase my teacher says we can't use a calculator on this problem? Just wanted to make sure since don't you have to do some substitution of "x" to find where it's increasing/decreasing?
 
  • #12
Think of the graph of x2. What is the value of the derivative when it changes from positive to negative (the slope of the tangent line at the vertex)? No calculator needed.
 
  • #13
Oh ok .. good, x squared is easier for me. :)

So the value of the derivative for x squared increases all the way through the graph (from left to right across the x axis). Derivative value at the origin is zero.

The derivative graph is basically a straight line going northeast by going north 2 spaces and east 1 space.
 
  • #14
Now how do you find where the derivative of xe1-x changes from positive to negative, or vice versa?
 
  • #15
When x is more than than one, because 1-x means that the chain part of the derivative is negative, while the other part (e^(1-x)) is negative ... wait, nevermind that's 2 negatives. Hm ... I'm not sure.
 
  • #16
When is the derivative of xe1-x equal to 0? One one side of that point (or points) the derivative will be positive and the other side will be negative. Think back to x2.
 
  • #17
The derivative of xe[tex]^{1-x}[/tex] is zero when x=1.

Ah, so the derivative of xe[tex]^{1-x}[/tex] is negative when x < 1 and positive when x > 1.
 
  • #18
Check the signs again for the derivative for x<1 and x>1.

For b, how do you think you would go about finding the range? Is there a maximum or minimum value for the function?
 
  • #19
Ack! Sorry, it's actually positive when x < 1 (I then took out my graphing calculator to prove this, just for confirmation).

For the range, I know the e graph doesn't have a negative y limit so the der. wouldn't have a POSITIVE y limit, right?

(Gotta go offline now, but be back later, thanks for the help so far!)
 
  • #20
Think about this:
Does f(x) go to positive infinity and negative infinity? If it doesn't go to one of them, say positive infinity for example, how do you find what the maximum value of f is? Think about the derivative...
 
  • #21
So, part b:

The range of f is y =< 1.

The thing is, I found this by my calculator, but apparently there's got to be a way to find this without calculator use.
 
  • #22
Wait, setting f(x) to equal zero or something should find the max value of f, I think.
 
Last edited:
  • #23
No, setting the derivative f'(x) equal to 0 will find a local max or min on the graph. Does y go to positive and/or negative infinity?
 
  • #24
Got this problem solved, thanks for your help, you got me on the right track. :)

(I already handed in this problem for my hw. Yeah, I'm rusty with this stuff and forgot that whatever makes the first derivative zero is the max or min and whatever makes the SECOND derivative finds the points of inflection,etc.)
 

1. What is the purpose of finding the intervals, ranges, and points of inflection for a function?

Finding the intervals, ranges, and points of inflection for a function allows us to understand and analyze the behavior of the function. It helps us identify where the function is increasing, decreasing, or constant, as well as any points where the direction of the function changes.

2. How do you determine the intervals where a function is increasing or decreasing?

To determine the intervals where a function is increasing or decreasing, we can take the derivative of the function and set it equal to 0. Then, we can solve for the x-values to find the critical points. Test points within each interval to determine if the function is increasing or decreasing.

3. What is a point of inflection?

A point of inflection is a point on a graph where the concavity changes from concave up to concave down, or vice versa. In other words, it is where the curvature of the graph changes.

4. How do you find the points of inflection for a function?

To find the points of inflection for a function, we take the second derivative of the function and set it equal to 0. Then, we solve for the x-values to find the possible points of inflection. We can also use the second derivative test to confirm if a point is a point of inflection.

5. Can a function have more than one point of inflection?

Yes, a function can have multiple points of inflection. This can happen if the concavity of the function changes multiple times within the domain of the function.

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