f(x,y) = exp(x)*[x - (y^3)/3 + y].
We are to locate and classify the stationary points of the surface given by the equation f(x,y) = 0.
I have tried answering this question many times now.. I get that there are no stationary points at f(x,y) = 0 (although f(x,y) has two stationary points.) Is this right? It is possible that have interpretted the question wrong... please have a look at what work i have done...
let fx = derivative with respect to x:
fx = exp(x) * [1 + x - (y^3)/3 + y];
fy = exp(x) * [1 - y^2]
at stationary point, fx = fy = 0. As exp(x) > 0 for all x, hence, it means that 1 + x - (y^3)/3 + y = 0 and 1 - y^2 = 0 which implies that y = +_ 1 (plus or minus 1)... from substition we get that x = -5/3 and -1/3. Hence two stationary points in f(x,y) are (-5/3,1) and (-1/3,-1).... but none satisfy f(x,y) = 0.... hence I think that f(x,y) = 0 has NO stationary points... what do u reckon?