Partial derivatives

1)
Partial derivatives
f x=−2sin 2 x=0=¿ x= π
2 ,− π
2 , 0
f y=2 ( y−1 ) =0=¿ y =1
The critical points are
( π
2 , 1) , ( −π
2 , 1 ) , ( 0,1 )
f xx =−4 cos 2 x
f yy=2
f xy =0
At ( π
2 , 1),
f xx =4
AC−B2 >0
Similarly at (−π
2 , 1 ),
f xx =4
AC−B2 >0
Similarly at (0, 1)
f xx =−4
AC−B2 <0
( π
2 , 1 ) , ( −π
2 , 1 ) arethe local minimum points
( 0,1 ) is a saddle point
2)
∇ . ⃗ f = ∂ f ( x , y , z)
∂ x i .i+ ∂ f ( x , y , z )
∂ y j . j+ ∂ f (x , y , z)
∂ z k . k
∇ . ⃗ f =x2 + x + ( 4 x+6 )=0=¿ x2+5 x +6=0
( x +5 ) ( x+1 ) =0=¿ x=−5∨−1
3)

∇ × ⃗ f = ( ∂
∂ x i+ ∂
∂ y j+ ∂
∂ z k ) ×(3 i+5 xz j+ y x2 k)
∇ × ⃗ f = ( x2−5 x ) i+ ( 0−2 xy ) j+ ( 5 z−0 ) k
∇ × ⃗ f = ( x2−5 x ) i−2 xy j+ 5 z k
( x2−x )=0=¿ x=0∨5
2 xy=¿ x∨ y=0
5 z=0=¿ z=0
The points where the curl are zero are
(0, y, 0) or (5, 0, 0) or (0, 0, 0)
4)
The x-axis represent the x values from −8 ≤ x ≤ 8 and the y axis represent the corresponding
f ( x )
f ( x ) = { 0−2< x< 0
2− x 0< x <2
Since it has a period of 4
f ( x ) = { 0−6< x ←4
2− x−4 < x←2
f ( x )= { 0−10< x ←8
2− x−8< x ←6
f ( x ) = { 0 2< x <4
2− x 4 < x <6
f ( x )= { 0 6< x <8
2− x 8< x <10
But since it is given that we can plot the graph between −8 ≤ x ≤ 8
So, the graph looks like,