If there is such a scalar function <em>f</em>, then



Integrate both sides of the first equation with respect to <em>x</em> :

Differentiate both sides with respect to <em>y</em> :


Integrate both sides with respect to <em>y</em> :

Plug this into the equation above with <em>f</em> , then differentiate both sides with respect to <em>z</em> :



Integrate both sides with respect to <em>z</em> :

So we end up with

Wei la dooblave hexagono 8 solutes x n o solutes = questionnairre
Hello,
y=5+3*cos (2(x-π/3))
The function is periodic with periode=2π.
-1<=cos (2(x-π/3))<=1
==>-1*3<=3*cos (2(x-π/3))<=3*1
==>5-3<=5+3cos(2(x-π/3))<=5+3
==>2<= y<=8
Answer: If I remember correctly, the answer should be <em>y=-2(3)^x. </em>