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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.

2 Answers 2

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

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    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 $(a,b)=(0,\pm i)$.2012-01-10
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$f_x(x,y)$ 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.