ECON 331
Solution Homework #3
2
4x
1. Consider the following function defined on I = [0, +∞), y = f (x) = x+3
.
(a) On I, f is a rational function and its denominator is nonzero. Hence, f is differentiable on I. Its derivative is:
f 0 (x) =
8x(x + 3) − 4x2
4x2 + 24x
4x(x + 6)
=
=
2
2
(x + 3)
(x + 3)
(x + 3)2
The denominator of f 0 , (x + 3)2 , is obviously strictly positive on I. The numerator,
4x(x + 6), is positive on I and only 0 at x = 0. We conclude that f is strictly increasing
on I.
(b) f is a one-to-one function on I because it is strictly increasing. Hence, f possesses
a unique inverse function f −1 . To find it, we need to solve the following equation in x:
4x2 − y(x + 3) = 0 ⇔ 4x2 − yx − 3y = 0
This is a quadratic equation. Its discriminant is: ∆ = y 2 + 48y ≥ 0 (because f has
positive values on I). There is at least one solution, potentially 2 if ∆ > 0:
p
p
y + y 2 + 48y
y − y 2 + 48y
x1 =
and x2 =
8
8
We know that the inverse function is unique, so only one of these can be the actual
inverse function of f . Recall that on I, x ≥ 0. x2 cannot be the inverse function of f
because it may lead to negative values of x:
y ≥ 0 ⇔ y 2 + 48y ≥ y 2 ≥ 0 ⇒ y −
p
y 2 + 48y ≤ 0
h
i
p
To conclude, the inverse function of f on I is f −1 (y) = 18 y + y 2 + 48y .
(c) f −1 is differentiable on I as a composite function of a polynomial function x2 + 48x
(positive on I) and the square-root function.
·
¸
1 1 1
1 1
2
−1/2
−1 0
(2x + 48)(x + 48x)
(f ) (x) = +
= + (x + 24)(x2 + 48x)−1/2
8 8 2
8 8
1
(d) Recall the expression for the firm’s average cost function:
r
C(y)
f −1 (y)
5 5 y + 48
AC(y) =
= 10
= +
y
y
4 4
y
This function is well-defined on (0, ∞). Moreover, it is a composite function of a rational
function and the square-root function: the rational function has a nonzero denominator
on (0, ∞) and is positive on (0, ∞). So it is differentiable on (0, ∞).
¶−1/2
y + 48
y − (y + 48)
×
y
y2
−5
p
=
<0
for any y > 0
2y y(y + 48)
5
AC (y) =
8
0
µ
We conclude that the average cost function is strictly decreasing on (0, ∞).
2. (a) R0 (p) = D(p) + pD0 (p). Roughly speaking, increasing the price by 1 has the
immediate effect of increasing the revenue by D(p). But, the higher price reduces demand (presumably D0 (p) is negative ie the demand decreases when the price increases)
and reduces the revenue by approximately (−pD0 (p)).
(b)
π(L) = F (f (L)) − w(L)L
⇒ π 0 (L) = F 0 (f (L))f 0 (L) − w0 (L)L − w(L)
⇒ π 00 (L) = F 00 (f (L))(f 0 (L))2 + F 0 (f (L))f 00 (L) − w00 (L)L − 2w0 (L)
3. Consider the function f (x) = ln(x + 1) with domain Df = [0, 10].
(a) The Taylor’s formula (around a = 0) with n = 3 is:
f (x) = f (0) + f 0 (0)x +
f 00 (0) 2 f (3) (0) 3 f (4) (c) 4
x +
x +
x
2
3!
4!
where 0 < c < x. After simplifications, we find:
ln(1 + x) = x −
x2 x3
x4
+
−
2
3
4(1 + c)4
2
(b) We want to provide an estimate of ln(1.01). We can use the formula found in (a)
and apply it for x = 0.01:
0.012 0.013
0.014
ln(1.01) = 0.01 −
+
−
2
3
4(1 + c)4
0.012 0.013
⇒ ln(1.01) ∼ 0.01 −
+
= 0.00995
2
3
where 0 < c < 0.01
(c) We need to bound the error associated to the approximation in (b) ie we need to
0.014
find an upper bound for | − 4(1+c)
4 |. We know that 0 < c < 0.01, so we have:
0 < c < 0.01 ⇔ 1 < 1 + c < 1.01
⇔ 1 < (1 + c)4 < 1.014
1
1
⇔
<
<1
1.014
(1 + c)4
0.014
0.014
0.014
⇔
<
<
= 2.5 × 10−9
4
4
4 × 1.01
4(1 + c)
4
To conclude, 0.0095 gives an approximation of ln(1.01) with an error less than 2.5×10−9 .
4. (a)
y0 =
(b)
−6xy
=3
3x2 − 5y 4
at (1,1)
(6y + 6xy 0 )(3x2 − 5y 4 ) − 6xy(6x − 20y 3 y 0 )
y =−
= −69
(3x2 − 5y 4 )2
00
at (1,1)
(c) We differentiate wrt to r:
f 00 (T )
dT
dT
f (T )
f (T )
= f (T ) + rf 0 (T ) ⇒
= 00
= 00
0
dr
dr
f (T ) − rf (T )
f (T ) − r2 f (T )
3