Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know title sounds weird but i had to translate it and that was best i could put. Anyhow i have the following function:

$$ f(x) = x\cdot e^{-x^{2}} $$

I have to find the following:
1. Intervals where function is falling and rising ?
2. Convexity intervals
3. Local and global extreems
4. Graph the function

I was sick and was not able to attend the last class so now I'm looking at the following function and questions not knowing where to find the following.

To solve ( 1. ) i assumed i need to find the derivative of the function which I did and I got $ f'(x) = 1 \cdot e^{-x^{2}} + (-2x^2 \cdot e^{-x^{2}}) = e^{-x^{2}} \cdot (-2x^2 + 1) $

Now that i got the 1st derivative of the function i wanted to check following two rules: Function is falling if $ x = f(x) ; x < x + 1 $ and rising $ x = f(x); x + 1 > x $

However I'm compleately lost here.

share|improve this question
1  
The function is falling if $f'(x) \lt 0$ and rising if $f'(x) \gt 0$. Since $e^{-x^2} \gt 0$, this changes when $2x^2=1$, i.e. when $x=\pm \sqrt{\frac12}$. –  Henry Dec 30 '12 at 2:16
    
Hints: Here are some notes you missed for convexity. Start with a plot –  Amzoti Dec 30 '12 at 2:17
    
It's driving me nuts, iv been at it for 4h nothing. –  miha Dec 30 '12 at 2:58
add comment

2 Answers

  1. If you know where $f(x)$ changes from falling to rising or vice versa (i.e. $f'(x) = 0$, then you can test $f'(x)$ around those critical points to see if it is rising or falling. If $f'(x) > 0$ then the slope is positive, so the function is rising, and if $f'(x) < 0$ then the slope is negative, so the function is falling.

    As given above, our critical points are when $-2x^2 = -1$, or $x = \pm\sqrt{\frac{1}{2}}$. Let's look at $x = \sqrt{\frac{1}{2}}$. If we plug in $x = 0$, then we get that $f'(0) = 1$. This tells us that the slope is positive between $-\sqrt{\frac{1}{2}}$ and $\sqrt{\frac{1}{2}}$. Hence the function is rising between those two values. What is it doing between $(-\infty, -\sqrt{\frac{1}{2}})$ and $(\sqrt{\frac{1}{2}}, \infty)$ then?

  2. Inflection points are when the concavity changes, just like how critical points are when the slope changes. Second derivatives tell you about concavity, just like first derivatives tell you about slope. Hence when $f''(x) = 0$, the concavity is changing, when $f''(x) > 0$, the graph is concave up or convex (U-shaped), and when $f''(x) < 0$, the graph is concave down or just concave (mountain-shaped). You can use the same strategy as in part (1) to determine concavity: look for where $f''(x) = 0$, then check values to the left and to the right of them to determine the concavity/convexity.

  3. Check what value $f(x)$ takes at your critical points and at your endpoints. The largest one will be your global maxima and the smallest one will be your global minima. Any critical points which aren't global maxima or minima are instead local maxima or minima.

  4. Based on parts 1, 2 and 3 you should have more than enough information to graph the function: you know when it is falling and rising, you know when it is concave and convex, and you know where the function has local and global maxima and minima.
share|improve this answer
    
I hope the OP can get the answer from our complete answers. –  B. S. Dec 30 '12 at 9:56
    
Thanks a lot mate this helps a lot. –  kellax Dec 30 '12 at 19:24
add comment

1) To have $f(x)$ increasing, if $f(x)$ be differentiable, we should have $f'(x)>0$. As you calculated $f'(x)$ so in this case $e^{-x^2}(1-2x^2)>0$. Note that $e^{-x^2}\neq 0$ so $$f'(x)>0\to 1-2x^2>0\to 2x^2<1\to x^2<1/2\\\to -\sqrt{2}/2<x<\sqrt{2}/{2}$$ This means that over $(\frac{\sqrt{-2}}{2},\frac{\sqrt{2}}{2})$, the function is increasing. Do the same for $f'(x)<0$. What intervals will we get?

3) Obviously $f'(x)$ is defined for all reals so find the critical points just among $x$ which make $f'(x)$ vanished. We found them before above during finding intervals. What they are? They are $x_1=\frac{\sqrt{-2}}{2}, x_2=\frac{\sqrt{2}}{2}$. Now after finding the decreasing intervals as I suggested above you will found out that; before $x_1$, the function $f(x)$ is decreasing and after that until we get $x_2$, the function is increasing, so $x_1$ can give us a relative minimum. The same deduction shows that $x_2$ makes a relative maximum for $f(x)$. Now consider the function as: $$f(x)=\frac{x}{e^{x^2}}$$ You see when $x\to\pm\infty$, $f(x)\to 0$. This means that $x_1$ and $x_2$ can be absolute extreme for $f(x)$ over its domain $\mathbb R$.

4) I used Maple with the following code:

plot(x*exp(-x^2),x=-5..5);

and got:

enter image description here

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.