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I'm looking at this function $f(x,y) = a + ab^2$

The derivates would be:

$df(a,b) \over da$$= 1+b^2$

$df(a,b) \over db$$= 2ab$

Now from what I understand there are no stationary points, but could someone please explain the logic reasoning behind that? How can I reach the conclusion that there are not stationary points?

I'm thinking in equation 2 it's not possible to reach any conclusion because $a$ could be $0$ and then $b$ could be anything and vice versa.

In Equation 1 I assume because there are only one variable, it can be determined.

Is this reasoning correct and is there anything else that could be said regarding this?

Thank you.

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2 Answers 2

up vote 1 down vote accepted

If $b$ is real, then $1+b^2$ can't be zero.

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Tack! Är det de definitiva svaret eller har mitt ovanstående någon sanning i sig? –  samuelf Jan 10 '12 at 17:44
Hej! :) Det beror på definitionsmängden. Frågor av den här typen brukar oftast uppstå i sammanhang där variablerna är reella (typ flervariabelanalyskurser), men det är ju inget som hindrar att man ställer denna fråga för samma funktion betraktad som funktion av två komplexa variabler, och då har den ju två (komplexa) stationära punkter $(a,b)=(0,\pm i)$. –  Hans Lundmark Jan 10 '12 at 18:12

$f_x(x,y)$ cannot be 0 if b is a real number . $\frac{dy}{dx}$=$\frac{-f_a(x,y)}{f_b(x,y)}$ so at any point it can't be 0 if $x,y\in\mathbb{R}$ and obviously if x or y=0 a vertical tangent line is resulted.

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