The First Derivative Test

[sammy87]

I have to use the first derivative test to find the local extrema and any absolute extrema.

y = xe^1/x

d/dx = (1)(e^1/x) + (e^1/x)(-1/x²)(x) (product rule) (-1/x² is the derivative of 1/x)
d/dx = e^1/x - (1/x)e^1/x

I set this equal to 0 and solved for x.

0 = e^1/x - (1/x)e^1/x
(1/x)e^1/x = e^1/x (divide both sides by e^1/x)
1/x = 1
1 = 1(x)
1 = x

The problem is that on the graph, -1 is also a local extrema. So why did it only come out to +1? :shock: Help!

 

[Unknown008]

I have to use the first derivative test to find the local extrema and any absolute extrema.

y = xe^1/x

d/dx = (1)(e^1/x) + (e^1/x)(-1/x²)(x) (product rule) (-1/x² is the derivative of 1/x)
d/dx = e^1/x - (1/x)e^1/x

I set this equal to 0 and solved for x.

0 = e^1/x - (1/x)e^1/x
(1/x)e^1/x = e^1/x (divide both sides by e^1/x)
1/x = 1
1 = 1(x)
1 = x

The problem is that on the graph, -1 is also a local extrema. So why did it only come out to +1? :shock: Help!

How did you find the local extrema at x = -1? Factoring says that there is a 'broken point' at x = 0.

$\displaystyle \displaystyle0 = e^{1/x} - \frac{1}{x}e^{1/x}$

$\displaystyle \displaystyle0 = e^{1/x}(1 - \frac{1}{x})$

Then, you have:

$\displaystyle \displaystyle 0 = 1 - \frac{1}{x}$ OR $\displaystyle \displaystyle 0 = e^{1/x}$

Which give:

$\displaystyle \displaystyle x = 1$ OR $\displaystyle \displaystyle -\infty = \frac{1}{x}$

 

[lookagain]

$\displaystyle \displaystyle0 = e^{1/x} - \frac{1}{x}e^{1/x}$

$\displaystyle \displaystyle0 = e^{1/x}(1 - \frac{1}{x})$

Then, you have:

$\displaystyle \displaystyle 0 = 1 - \frac{1}{x}$

which gives:


$\displaystyle \displaystyle x = 1 \ \ OR$


$\displaystyle 0 = e^{1/x}$


which gives:


$\displaystyle no \ solution$


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Why type $\displaystyle " \ -\infty \ = \ \frac{1}{x} \ ?"$


Did you mean $\displaystyle " \ -\infty \ = \ x \ ?"$


But then what about $\displaystyle " \ \infty \ = \ x \ ?"$


You wouldn't type that. You would type something along
the lines of "no solution" for that part of the problem
or "discard/can't use this."


There is one critical number, and it is x = 1.


Also, separate your equations and their respective solutions by horizontal
spacing of lines.