Math 3220-1 Midterm 2, November 18, 2015
Solutions
Problem 1 (20 points). Let f : R −→ R be a differentiable function
on R. Define a function F : R2 −→ R by F (x, y) = f (x + y). Show
that
∂F
∂F
=
.
∂x
∂y
Solution: By the chain rule have
∂F
∂F
= f 0 (x + y) and
= f 0 (x + y)
∂x
∂y
for any x, y ∈ R2 . This implies that
∂F
∂F
= f 0 (x + y) =
.
∂x
∂y
Problem 2 (20 points). Prove that if f is a real-valued function defined
on an open interval of R containing a and if S is an affine function such
that f (a) = S(a) and
f (a + h) − S(a + h)
lim
= 0,
h→0
h
then S(a + h) = f (a) + f 0 (a)h.
Solution: The affine function S must have the form S(a + h) =
S(a) + ah = f (a) + Ah for some number A ∈ R. We then have
f (a + h) − S(a + h)
f (a + h) − f (a) − Ah
= lim
h→0
h→0
h
h
f (a + h) − f (a)
− A.
= lim
h→0
h
This implies that
f (a + h) − f (a)
lim
= A,
h→0
h
i.e., A = f 0 (a).
0 = lim
Problem 3 (20 points). Find all points of local maximum, local minimum and all saddle points of the function
f (x, y) = x3 + x2 + y 2 + 2xy − 3x + 1.
1
2
Solution: The partial derivatives of f are
∂f
= 3x2 + 2x + 2y − 3 and
∂x
∂f
= 2y + 2x.
∂y
Hence, (x, y) is a stationary point of f if
3x2 + 2x + 2y − 3 = 0
and
2y + 2x = 0.
From the second equation we see that y = −x. Substituting this in the
first equation we get
0 = 3x2 + 2x − 2x − 3 = 3x2 − 3.
Hence x2 = 1. It follows that x = ±1, and (1, −1) and (−1, 1) are
stationary points.
The matrix of second derivatives
" 2
# ∂ f
∂2f
6x + 2 2
∂x2
∂x∂y
=
.
∂2f
∂f
2
2
2
∂x∂y
∂y
Therefore, in the first stationary
8
2
point (1, −1) it is equal to
2
.
2
The determinant is 16 − 4 = 12. Hence, the point (1, −1) is a local
minimum.
In the second stationary point (−1, 1) it is equal to
−4 2
.
2 2
The determinant is −8 − 4 = −12. Hence, the point (−1, 1) is a saddle
point.
Problem 4 (20 points). Let F, G : R2 −→ R2 be the functions given
by F (x, y) = (x2 + y 2 , xy) and G(x, y) = (x + y, x − y). Using the chain
rule find the differential dH of the composition H = F ◦ G. Check your
result by direct calculation!
Solution: The differential dF of F (x, y) = (F1 (x, y), F2 (x, y)) is
given by the matrix
"
# ∂F1
∂F1
2x
2y
∂x
∂y
dF (x, y) = ∂F
=
.
∂F2
2
y x
∂x
∂y
3
The differential dG of G = (G1 (x, y), G2 (x, y)) is given by the matrix
"
# ∂G1
∂G1
1 1
∂x
∂y
dG(x, y) = ∂G2 ∂G2 =
.
1 −1
∂x
∂y
Hence, we have
dH(x, y) = dF (G(x, y))dG(x, y)
2(x + y) 2(x − y) 1 1
4x 4y
=
=
.
x−y
x+y
1 −1
2x −2y
On the other hand, we have
H(x, y) = F (G(x, y)) = F (x + y, x − y)
= ((x + y)2 + (x − y)2 , (x + y)(x − y)) = (2x2 + 2y 2 , x2 − y 2 ).
Hence, by direct calculation, we have that the differential dH of H(x, y) =
(H1 (x, y), H2 (x, y)) is given by
"
# ∂H1
∂H1
4x 4y
∂x
∂y
dH(x, y) = ∂H2 ∂H2 =
.
2x −2y
∂x
∂y
Problem 5 (20 points). Let F (x, y) = (x − y, x2 + y 2 ) be a function
from R2 into R2 . Prove that it has a local inverse F −1 near the point
(0, 1). Find the differential dF −1 (−1, 1)!
Solution:
Put F (x, y) = (F1 (x, y), F2 (x, y)) with F1 (x, y) = x−y and F2 (x, y) =
x2 + y 2 . The differential dF of F is given by the matrix
"
# ∂F1
∂F1
1
−1
∂x
∂y
=
.
∂F2
∂F2
2x 2y
∂x
∂y
Hence, we have
1 −1
dF (0, 1) =
.
0 2
This matrix is invertible and F has a local inverse at (0, 1). Since
F (0, 1) = (−1, 1), the differential of F −1 is the inverse of that matrix,
i.e.,
1 2 1
−1
dF (−1, 1) =
.
2 0 1