Question

Determine the nature of the stationary value

X = t³ - 3t + ty²

Answer 1

$X_t=3t^2-3+y^2=0$

$X_y=2yt=0$

stationary points:

$t=0,y=0$

$y=0,t=\pm 1$

$t=0,y=\pm \sqrt 3$

$D=X_{tt}X_{yy}-(X_{ty})^2$

$X_{tt}=6t,X_{yy}=2t,X_{ty}=2y$

at point $y=0,t= -1$ :

$D=12>0,X_{tt}=-6<0,X_{yy}=-2<0$

local maximum

at point $y=0,t= 1$ :

$D=12>0,X_{tt}=6>0,X_{yy}=2>0$

local minimum

at points $t=0,y=\pm \sqrt 3$ :

$D=-12<0$

saddle points

at point $y=0,t= 0$ :

$D=0$

$X_t$ does not change sign ($X_t<0$ ) near this point, so this is not local maximum, not local minimum and not saddle point