# Find and sort all of the extremum of $f(x,y) = |3x| + x^4y^2$

Find and sort all of the extremum of $f(x,y) = |3x| + x^4y^2$

not quite sure what to do with the absolute value of $|3x|$ , should i separate it to two options? , one for $3x>0$ and the other option for $3x<0$ ?

• from $f(x,y) = |3x| + x^4y^2= 3|x| + |x|^4|y|^2$ I think you can discuss with $x\geq0$ and $y\geq0$ in $f(x,y) = 3x + x^4y^2$. – Nosrati Sep 14 at 13:19
• how do i differentiate it? $f_x = 3+4x^3y^2$ ? am i supposed to just ignore the absolute value? – Maor Rocky Sep 14 at 13:24
• Its minimum value is $0$, all along the $y$-axis. – Empy2 Sep 14 at 13:35

from $f(x,y) = |3x| + x^4y^2= 3|x| + |x|^4|y|^2$ I think you can discuss with $x\geq0$ and $y\geq0$ in $f(x,y) = 3x + x^4y^2$.
$f_x=3+4x^3y^2=0$
$f_y=2x^4y=0$
with the second $x=0$ or $y=0$ which both are impossible as conflict with the first, remains $$f(x,y) = |3x| + x^4y^2= 3|x| + |x|^4|y|^2\geq0=f(0,0)$$ specifies the only minimum point in the origin.
Hint: it is clear that $$3|x|+x^4y^2\geq 0$$ so the Minimum is given by $$(0,0)$$