Natural Exponential & Logarithmic Function

[Lime]

The question is to find where the function is increasing and decreasing.

The function is f(x) = x^2/e^x
The derivative of that is f'(x) = xe^-x (2 - x)

I get that. What I don't is how the solution book determines from this, that "f'(x) = 0 <--> x = 0 or 2"

1) Why does f prime equal 0?
2) Why does x equal 0?
3) Does x equal 2 because it will put a 0 inside the bracket above?
4) What does the double pointed arrow indicate?

 

[tkhunny]

There's a funcdamental theorem around here somewhere.

Continuous things have to go through zero (0) to switch from positive to negative. If you find whre the expression is zero, you define segments to investigate.

x*[e^(-x)]*(2-x) = 0

e^(-x) is never zero. How about the other two pieces?

The two-pronged arrow means "if and only if".

 

[soroban]

Hello, Lime!

Find where the function is increasing or decreasing: \(\displaystyle \L\,f(x)\:=\:\frac{x^2}{e^x}\)

As you said, the derivative is: \(\displaystyle \L\,f '(x) \:= \:\frac{x(2\,-\,x)}{e^x}\)
. . But evidently you don't understand what a derivative is ... or what it's for.

The derivative of a function gives us the slope of the graph.

And you should know already that:
. . if the slope is positive, the function is increasing;
. . if the slope is negative, the function is decreasing.

Very well, exactly where is the derivative positive?
. . Do we plug in dozens of numbers and test them? . . . No!

We find where the derivative is zero (neither increasing nor decreasing)
. . and test the slope around those values.

We set the derivative equal to zero: \(\displaystyle \L\,f'(x)\;=\;\frac{x(2\,-\,x)}{e^x}\;=\;0\)
. . and solve for \(\displaystyle x:\;\;x\,=\,0,\;2\)


These values divide the number line into three intervals.

. . \(\displaystyle \,\underbrace{\;- - - \,}+\underbrace{\, - - - - \,}+\underbrace{\, - - -\;}\)
. . . . . . . .\(\displaystyle 0\) . . . . . .\(\displaystyle 2\)


In \(\displaystyle (-\infty,\,0),\) try \(\displaystyle x\,=\,-1\)
. . We have: \(\displaystyle \,f'(-1)\:=\:\frac{(-1)(3)}{e^{-1}} \:=\:[-]\) . . . negative: decreasing \(\displaystyle \searrow\)

In \(\displaystyle (0,\,2(,\) try \(\displaystyle x\,=\,1\)
. . We have: \(\displaystyle \,f'(1)\:=\:\frac{1(1)}{e}\:=\:[+]\) . . . positive: increasing \(\displaystyle \nearrow\)

In \(\displaystyle (2,\,\infty),\) try \(\displaystyle x\,=\,3\)
. . We have: \(\displaystyle \,f'(3)\:=\:\frac{3(-1)}{e^3} \:=\:[-]\) . . . negative: decreasing \(\displaystyle \searrow\)


So we have:

. . . . \(\displaystyle \searrow\) . . . . . \(\displaystyle \nearrow\) . . . . \(\displaystyle \searrow\)
. . \(\displaystyle \,\underbrace{\;- - - \,}+\underbrace{\, - - - - \,}+\underbrace{\, - - -\;}\)
. . . . . . . .\(\displaystyle 0\) . . . . . .\(\displaystyle 2\)

Now it's obvious, isn;t it?

Decreasing on \(\displaystyle (-\infty,\,0)\) and \(\displaystyle (2,\,\infty)\)
Increasing on \(\displaystyle (0,\,2)\)