MATHEMATICS FOR ECONOMICS – 2014/2015
Sheet 3. Derivatives
3-1. Find the points where the following functions have an horizontal tangent line.
3
2
a) f (x) = x
c) f (x) = x + sin x
√ + 1 b) f (x) = 1/x
x
d) f (x) = x − 1 e) f (x) = e − x f) f (x) = sin x + cos x.
Sol. The tangent line is horizontal at points c where f 0 (c) = 0.
a) x = 0; b) At no point; c) At points c = (2k + 1)π, k integer; d) At no point; e) At c = 0; f) At points
c = (k + 1/4)π, k integer.
3-2. (*) Find the derivatives of the following functions:√
x x2 − 1
a) f (x) = sin x + tan 3x sin 2x
b) f (x) =
2x + 6
3/2
c) f (x) = 4x cos 2x
d) f (x) = 5x ln(8x + sin 2x) + etan 5x .
√
e) f (x) = xsin x
f) g(x) = ( x)x .
Sol. a) cos x + 3(1 + 3 tan2 3x) sin 2x + 2 tan 3x cos 2x;
2
x3 +6x√
−3
b) 2(x+3)
;
2 x2 −1
√
c) 6 x cos 2x − 8x3/2 sin 2x;
4+cos 2x
2
tan 5x
d) 5 ln (8x + sin 2x) + 10x 8x+sin
;
2x + 5(1 + tan 5x)e
sin x
sin x
e) x
(ln x cos x + x );
√ x
f) 12 ( x) (1 + ln x).
3-3. (*) The functions f and g are related by f (x) = 2[ln(1 + g 2 (x))]2 . Using that g(1) = g 0 (1) = −1,
evaluate f 0 (1).
Sol. f 0 (1) = 4 ln 2.
3-4. Sean f (x) = ln(1 + x2 ) y g(x) = e2x + e3x . Calcula (f ◦ g)0 (0) y (g ◦ f )0 (0).
Sol. (f ◦ g)0 (0) = 4 and (g ◦ f )0 (0) = 0.
3-5. (*) Find the derivative of the following functions, pointing out the points where they are not differentiable.

2
si x ≤ −2
 1/|x|
x − 1 si x ≤ 0
2
(x + 2)
si − 2 < x ≤ 0
a) f (x) =
b) g(x) =
0
si x > 0

3 + sin (x + π2 ) si x > 0
Sol. a) f is derivable in R − {0} but is not at 0 since f is not continuous at 0. Moreover, f 0 (x) = 2x if x < 0,
f 0 (x) = 0 if x > 0;
b) g is derivable in R − {−2, 0} but it is not at −2 since it is not continuous at this point, nor at 0;
3-6. (*) Find constants a and b such that the function f (x) =
3x + 2
si x ≥ 1
is differentiable.
ax2 + bx − 1 si x < 1
Sol. a = −3 and b = 9;
3-7. (*) Let f (x) = x3 − 3x + 3, where f : [−3, 2] → R. Find the global extremum of f .
Sol. −3 global minimum and −1, 2 global maximum;
1
2
√
3
3-8. Let f (x) = −2x + 3 x2 , where f : [−1/8, 27/8] → R. Find the global extremum of f .
Sol. 0, 27/8 global minimum and −1/8, 1 global maximum;
3-9. (*) Find the local extremum of the following functions:
a) f (x) = x4 − 4x3 + 2; b) g(x) = x2 (2 − x)3 ; c) h(x) = sin x − x;
√
x
d)j(x) = x + x4 ;
e) k(x) = x2 + 1;
f) l(x) = x−1
.
Sol. a) 3 local minimum;
b) 0 local minimum and 4/5 local maximum;
c) No local extremum.
d) −2 local maximum and 2 local minimum;
e) 0 local (and global) minimum;
f) No local extremum.
3-10. (*) Find the following limits using L’Hospital rule:
a) limx→∞
d) limx→0
2x+1
;
4x2 +x
ex −(1−x)
;
x
b) limx→0
e) limx→∞
ln (sin αx)
ln (sin βx) ;
ln
√x ;
x
√
c) limx→1
2−x−x
x−1
;
f) limx→0 (1 + x)1/x .
Sol. a) 0; b) 1; c) −3/2; d) 2; e) 0; f) e.
3-11. (*) Using the linear approximation f (c √
+ h) ≈ f (c) + hf 0 (c) and using the appropriated values of c,
h, find the approximated values of (a) 3 1.01 and (b) sin 31o .
Sol. (a) 1.0033; (b) 0.5151.
3-12. (*) Find the intervals of monotonicity of the functions (a) f (x) =
x > 0, where m, n are positive integers.
2x
1+x2
and (b) g(x) = xn e−mx , for
Sol. (a) Increasing in (−1, 1), decreasing in (−∞, −1) ∪ (1, ∞);
(b) The derivative is g 0 (x) = e−mx xn−1 (n − mx). Since x 6= 0 and the exponential never vanishes, g is
increasing in (0, n/m) and decreasing in (n/m, ∞).
3-13. (*) Find the intervals of concavity/convexity, as well as the inflection points of the functions (a)
f (x) = 1 + x2 − 21 x4 and (b) g(x) = xn e−mx , for x > 0, where m, n are positive integers.
√
√
√
√
Sol. (a) Concave in (−∞, −1/ 3) and in (1/ 3, ∞), convex in (−1/ 3, 1/ 3);
(b) The second derivative is g 00 (x) = e−mx xn−2 (n(n − 1) − 2mnx + m2 x2 ). Since e−mx xn−2 > 0, the sign of
g 00 depends on the sign
of the convex parabola n(n − 1) − 2mnx + m2 x2 (note that m2 > 0). The zeroes of
√
√
n± n
the parabola are m ≥ 0, thus g is convex in the intervals (0, (n − n)/m) (n > 1, otherwise this interval
√
√
√
is empty) and ((n + n)/m, ∞), and it is concave in ((n − n)/m, (n + n)/m).
3-14. (*) A monopolist produces a good in quantity x ≥ 0. The total cost of producing x units is given
by the function C(x). The market unitary price is given by the function p(x) (inverse demand
function). Assume that both functions are differentiable. Assume that the monopolist seeks to
maximize profits.
(a) Calculate the profit function.
(b) Show the quantity of the good that maximizes profits, if exists and positive, is where the
marginal cost equals marginal revenue.
(c) Write down sufficient conditions so that a critical point of the profit function is a local maximum. Which conditions guaranty that the critical point is a global maximum?
(d) Lat x1 > 0 (x2 > 0) be such that the marginal revenue is bigger (smaller) than the marginal
cost. What is worth doing, to rise, to diminish or to maintain production?
(e) Assume that C(x) = x2 − 10x − α, where α > 0 and p(x) = 100 − x2 . Solve the monopolist’s
problem and calculate the maximum profits when α = 200.
3
(f) The average cost is Cm (x) = C(x)
x . Use the data of part above to find the fixed cost α such
that the minimum of Cm and the maximizer of profits coincide
Sol. (a) B(x) = xp(x) − C(x), where I(x) = xp(x) is the revenue;
(b) If x0 exists, then B 0 (x0 ) = 0 or I 0 (x0 ) = C 0 (x0 );
(c) If x0 is critical point, then B 00 (x0 ) < 0 is a sufficient condition, that is, I 00 (x0 ) < C 00 (x0 ). B concave
implies it is gobal;
(d) Marginal cost greater than marginal revenue means that the monopolist has to reduce production to
increase profits;
(e) B ios concave, x0 = 30 and B(x0 ) = 1550 u.m.:
√
0
(d) Cm (x) = x + 10 + αx−1 . Cm
(x) = 1 − αx−2 . Thus, the minimum of Cm is xm = α. Then x0 = xm iff
α = 900.