Math Review
Solutions to Homework #5
1.
a) f(x) = 2 if x  0, and = x2 if x > 0. Note that f(0) = 2 and f(0 + ) = 0 for  > 0 but “infinitely
small”. It follows that, for any r > 0, there always exists an x  Br(0) such that ||f(x) – f(0)|| = 2.
Thus, there does not exist an r > 0 such that x  Br(0) implies that ||f(x) – f(0)|| < e for all e > 0.
This implies that f is not continuous at x = 0. Thus, f is not continuous on R. Not being
continuous at x = 0, f is not differentiable at x = 0. Thus, f  C1, and f  C2.
b) f(x) = 2 x if x  0, and = x2 if x > 0. Clearly, f is continuous and differentiable on R\{0}. Note
that f(0) = 0 and f(0 + ) = 0 for  “infinitely small”. It follows that there always exist an r > 0
such that x  Br(0) implies that ||f(x) – f(0)|| < e for all e > 0. Thus, f is continuous at x = 0.
We have
Df(x) = 2,
if x < 0
= 2 x, if x > 0.
However, limx0 {[f(x) – 0 – A (x – 0)]/x} = limx0 {[f(x) – A x]/x}
= limx0 [f(x)/x  A] = 2 – A if x < 0
= – A if x > 0.
Since we cannot have both (2-A) = 0 and A = 0, it follows that f is not differentiable at x = 0.
Thus, f  C1, and f  C2.
c) f(x) = 2 x2 if x  0, and = x2 if x > 0. Clearly, f is continuous and differentiable on R\{0}. Note
that f(0) = 0 and f(0 + ) = 0 for  “infinitely small”. It follows that there always exist an r > 0
such that x  Br(0) implies that ||f(x) – f(0)|| < e for all e > 0. Thus, f is continuous at x = 0.
We have
Df(x) = 4 x, if x < 0
= 2 x, if x > 0.
In addition, limx0 {[f(x) – 0 – A (x – 0)]/x} = limx0 {[f(x) – A x]/x}
= limx0 [f(x)/x  A] = – A if x < 0
= – A if x > 0.
It follows that f is differentiable at x = 0, with A = Df(0) = 0 as derivative. Thus, f is
differentiable on R.
Clearly, Df is continuous and differentiable on R\{0}. Note that Df(0) = 0 and Df(0 + ) = 0 for 
“infinitely small”. It follows that there always exist an r > 0 such that x  Br(0) implies that
||Df(x) – Df(0)|| < e for all e > 0. This implies that Df is continuous at x = 0, and therefore
continuous on R. Thus, f  C1.
We have
D2f(x) = 4,
if x < 0
= 2,
if x > 0.
In addition, limx0 {[Df(x) – 0 – A (x – 0)]/x} = limx0 {[Df(x) – A x]/x}
= limx0 [Df(x)/x  A] = 4 – A if x < 0
= 2 – A if x > 0.
Since we cannot have both (4 – A) = 0 and (2 – A) = 0, it follows that Df is not differentiable at x
= 0. Thus, Df is not differentiable on R, and f  C2.
d) f(x) = 2 x3 if x  0, and = x4 if x > 0. Clearly, f is continuous and differentiable on R\{0}. Note
that f(0) = 0 and f(0 + ) = 0 for  “infinitely small”. It follows that there always exist an r > 0
such that x  Br(0) implies that ||f(x) – f(0)|| < e for all e > 0. Thus, f is continuous at x = 0.
We have
2
= 6 x2, if x < 0
= 4 x3, if x > 0.
In addition, limx0 {[f(x) – 0 – A (x – 0)]/x} = limx0 {[f(x) – A x]/x}
= limx0 [f(x)/x  A] = – A if x < 0
= – A if x > 0.
It follows that f is differentiable at x = 0, with A = Df(0) = 0 as derivative. Thus, f is
differentiable on R.
Clearly, Df is continuous and differentiable on R\{0}. Note that Df(0) = 0 and Df(0 + ) = 0 for 
“infinitely small”. It follows that there always exist an r > 0 such that x  Br(0) implies that
||Df(x) – Df(0)|| < e for all e > 0. This implies that Df is continuous at x = 0, and therefore
continuous on R. Thus, f  C1.
We have
D2f(x) = 12 x, if x < 0
= 12 x2, if x > 0.
In addition, limx0 {[Df(x) – 0 – A (x – 0)]/x} = limx0 {[Df(x) – A x]/x}
= limx0 [Df(x)/x  A] = – A if x < 0
= – A if x > 0.
It follows that Df is differentiable at x = 0, with D2f(0) = A = 0. Thus, Df is differentiable on R.
Clearly, D2f is continuous and differentiable on R\{0}. Note that D2f(0) = 0 and Df2(0 + ) = 0 for
 “infinitely small”. It follows that there always exist an r > 0 such that x  Br(0) implies that
||f(x) – f(0)|| < e for all e > 0. This implies that D2f is continuous at x = 0. Thus, D2f is continuous
on R, and f  C2.
Df(x)
2.
(a) For x0 = 0, limxxo (x2 sin(1/x2) = 0 = f(x0) (since x2  0 and -1  sin(1/x2)  1). For x0  0,
limxxo(x2 sin(1/x2)) = limxxo(x2) limxxo sin(1/x2) = x02 sin(1/x02) = f(x0). So f is continuous on
R.
(b) For x  0, f is continuously differentiable, since
Df(x) = 2x sin(1/x2) - (2x3/x4) cos(1/x2)
= 2x sin(1/x2) - (2/x) cos(1/x2)
is continuous at x  0. Still, we need to show that f is differentiable at x = 0. From the definition
of the derivative, at x=0,
Df(x) = limx0 [(f(x)-f(0))/(x-0)]
= limx0 [x2 sin(1/x2)/x]
= limx0 (x sin(1/x2)) = 0, since |sin(1/x2)|  1.
So, Df(0) = 0 exists.
(c) limx0 Df(x) = limx0[2x sin(1/x2) - (2/x) cos(1/x2)]
= limx0[2x sin(1/x2)] - limx0[(2/x) cos(1/x2).
This limit does not exist at x = 0, since -1  cos(1/x2)  1, but (2/x) as x0.
Thus, Df(x) is not continuous at x = 0.
3.
(a)
f
(x, y) =
x
y
( x2 + y2 )-1/2 x2 y
, (x, y)  (0, 0)
2
( x2 + y2 )
x2 + y
f
f(x,0) - f(0,0)
0-0
(0,0) = lim [
] = lim [
] = 0 , (x, y) = (0, 0)
x 0
x 0
x
x
x
-
3
f
(x, y) =
y
x
( x2 + y2 )-1/2 xy2
, (x, y)  (0, 0)
2
( x2 + y2 )
x2 + y
f(0, y) - f(0,0)
f
0-0
(0,0) = lim [
] = lim [
] = 0 (x, y) = (0, 0).
y0
y 0
y
y
y
So the partial derivatives of f exist everywhere.
(b) Notice that:
f(x, y) - f(0, y)
y
f
(0, y) = lim [
] = lim [
] = 1 or -1, for x = 0, y  0,
2
x 0
x0
x
x
x2 + y
and
f(x, y) - f(x,0)
f
x
(x,0) = lim [
] = lim [
] = 1 or -1, for y = 0, x  0.
2
y0
y 0
y
y
x2 + y
But:
f
f
(0,0)  lim y0
(0, y) =  1
x
x
f
f
0 = (0,0)  lim x 0
(x,0) =  1
y
y
0=
So Df is not continuous at (x, y) = (0, 0).
(c) Assume that f is differentiable at (x, y) = (0, 0).
f f
Df(0,0) = ( ,
)(0,0) = (0, 0).
x y

 x 
 ||f(x, y) - f(0,0) - Df(0,0)   || 
 y 
= 0
lim 

(x,y)  (0,0)
||(x, y) - (0,0)||




Choose x = y = a. This implies

a  
 ||f(a, a) - f(0,0) - Df(0,0)   || 
a  
= 0,
lim 

a0
||(a, a)||




or
a2 / 2 a2
] = lim (1 / 2) = 0,
lim [
a0
a0
2 a2
so we have a contradiction. Thus, f is not differentiable at (x, y) = (0, 0).
4.
f(x) = a + b x + c x2
||a + by + cy2 - (a + bx + cx2)||
||f(y) - f(x)||
f (x) = lim [
] = lim [
]
y x
y x
||y - x||
||y - x||
4
||b(y - x) + c(y2 - x2)||
= lim [
]
y x
||y - x||
= lim [b + c (
y x
5.
y2 - x2
)] = lim (b + c (y + x)) = b + 2 c x .
y x
y-x
0 
2
1 3/2 
(a) D2f(x) = 
0  4 x 2 
2 0 
1
D2f(x0) = 
at x0 = (1, 1)
0  4 
The eigenvalues are 2 and -1/4, so the matrix is indefinite.
 1 -3/2 1/2 1 -1/2 -1/2 
x1 x 2 
 x 1 x 2
4
(b) D2f(x) =  4

1
1
-3/2
 x 1-1/2 x -1/2
 x 1/2
2
1 x2 
4
4

1 
 1


4  at x0 = (1, 1).
D2f(x0) =  4
1
1 

 
4 
 4
The eigenvalues are 0 and -1/2, so the matrix is negative semi-definite.
 2 x 22
4 x1 x2 
(c) D2f(x) = 

2 x 12 
4 x 1 x 2
2 4 
D2f(x0) = 
 at x0 = (1, 1).
4 2 
The eigenvalues are -2 and 6, so the matrix is indefinite.
 1 -3/2

- x1
0 

(d) D2f(x) =  4

1
 0
- x -3/2
2 
4


 1

0
D2f(x0) =  4
at x0 = (1, 1).
1
0 - 
4

The eigenvalues are (-1/4, -1/4), so the matrix is negative definite.
0 1
(e) D2f(x) = 
 for all x.
 1 0
The eigenvalues are (1, -1), so the matrix is indefinite.
5
0 0 0
(f) D2f(x) = 0 0 0 for all x.
0 0 0
The eigenvalues are (0, 0, 0), so the matrix is both positive semi-definite and negative semidefinite.