Universidad Carlos III de Madrid

Transcription

1 Universidad Carlos III de Madrid Exercise Total Points Department de Economics Mathematicas I Final Exam January 0th 07 Exam time: hours. LAST NAME: FIRST NAME: ID: DEGREE: GROUP: () Consider the unction (x) = ln( x ) + 3. Then: (a) Draw the graph o the unction, obtaining irstly its domain, symmetries and intersecting points with both axis, intervals where (x) increases and decreases, asymptotes and range o (x). (b) Consider the new unction (x) = (x), deined only on the interval where that unction is increasing. Draw the graphs o (x), its inverse and their corresponding tangent lines at the point x = 3, using the concavity and / or convexity o (x) and (x). Hint: you can ind the analytical expression o (x), but it is not necessary. 0.6 points part a); 0. points part b) a) The domain o the unction is {x : x > } = (, ) (, ). On the other hand, the unction is even, so its graph will be symmetric with respect to the vertical axis. For that reason, it is enough to study it on the interval (, ). The graph has no intersection points with the vertical axis, because x = 0 is not in the domain o the unction. Regarding the horizontal axis, i x > its intersection point will be x: ln(x ) = 3 x = e 3 x = + e 3. And, by symmetry, i x <, the intersection point will be x = e 3. Moreover, as (x) = (i x > ), you can deduce that is increasing on (, ) and, by x symmetry, decreasing on (, ). As the unction is continuous on its domain, it is only necessary to study the possible vertical asymptotes in + and in : lim x +(x) = ln(0+ ) + 3 = + 3 = ; so the unction has a vertical asymptote in x = + and, by symmetry, in x =. And, about the asymptotes in ±, as: (x) lim (x) = ; lim = 0; so the unction does not have any horizontal or oblique asymptotes x in and, by symmetry, neither in. For that reason, as lim x +(x) =, lim (x) = and the unction is continuous on the interval (, ), by the intermediate values theorem the image will be (, ). Conclusion: the graph o will have an appearance, approximately, as the irst drawing: 5 - (5,3+ln 3) 3 (3,3) 3 (3,3) y=x 3 5 b) I y = (x) >, (y) = ln(y ) + 3 = x ln(y ) = x 3

2 y = e x 3 y = (x) = + ex 3. But the problem can be solved without computing the analytical expression o the inverse unction. As (3) = ln(3 ) + 3 = 3, (3) = 3. Analogously, as (x) = x = (3) = = ( ) (3) =, so both unctions have the same tangent line y 3 = x 3 (i.e., y = x) in x = 3. Finally, as is a concave unction, because (x) = (x ) < 0, and increasing, it can be deduced that (x) is both convex and increasing. For those reasons, the graph o will be drawn under its tangent line y = x, and the graph o will remain above the same line, and both graphs will intersect at the point (3, 3). The relative position o both graphs, with respect to their common tangent line at the point x = 3, will be, near the point (3, 3), approximately, as the second drawing.

3 () Let y = (x) be an implicit unction deined by the equation ye x y x + x =, in a neighbourhood o the point x = 0, y =. Then: (a) Find the tangent line and the Taylor polynomial o degree o the unction centered at a = 0. (b) Use the tangent line and the Taylor polynomial to obtain an approximation o the value o (0, ). Can you justiy i any o these approximations are by deault or by excess? 0.5 points part a); 0.5 points part b) a) First o all, we compute the irst derivative o the unction: ye x + y e x yy x y + = 0 substituting x = 0, y(0) = in the previous equation, you obtain: y (0) = (0) =. Analogously, we compute the second derivative o the unction: (y + y + y)e x (y ) x yy x yy = 0 substituting x = 0, y(0) =, y (0) = in this last equation, you obtain: y (0) = (0) = 3 So, the equation o the tangent line will be: y = P (x) = + ( )(x 0), i.e. y = x. And the equation o the Taylor polynomial will be: y = P (x) = x 3 x. b) First order approximation: (0, ) P (0, ) = 0.9. Second order approximation: (0, ) P (0, ) = As the unction is concave near the point x = 0, because (0) < 0, the approximation by the tangent line is by excess. We cannot claim anything about the second order approximation.

4 (3) Let C(x) = 7 + 9x + x be the cost unction and p(x) = 8 ax be the inverse demand unction o a monopolistic irm, being x 0 the number o units produced o certain good and a > 0. Then: (a) Find the production that maximizes the proit. (b) Find the minimum average cost. Hint or b: irst o all, ind the production that minimizes the average cost. 0.5 points part a); 0.5 points part b) a) First o all, we calculate the proit unction: B(x) = (8 ax)x (7 + 9x + x ) = ( a )x + 7x 7 Secondly, we compute the irst and second order derivative o B : B (x) = ( a )x + 7; B (x) = ( a ) < 0 so we see that B has an unique critical point when x = 36 and, as B is a concave unction, (a + ) this critical point is the unique global maximizer. b) The average cost unction is C m (x) = C(x) = 7 x x x. I we compute the irst and second order derivatives o this unction: C m(x) = 7 x + ; C m(x) = 7 x 3 > 0 we observe that x = 6 is the only critical point o the unction C m (x) and, as this unction is convex, that critical point is the only global minimizer. Thereore, the production that minimizes the average cost is: x = 6. Finally, substituting in the average cost unction, the minimum average cost will be: C m (6) = = = 33 6

5 { ax + 3 i x < () Let the unction (x) = x and let us consider the interval [, 3] + ax + b i x. Then: (a) Find a and b in order that (x) satisies the hypothesis (or initial conditions) o Lagrange s theorem (or mean value theorem) in that interval. (b) For those values a, b ind the intermediate value or values c in such a way that the thesis (or conclusion) o this theorem is satisied. Hint or both parts: state the mean value theorem. 0.6 points part a); 0. points part b) a) We need to set the continuity and derivability at x =. For that reason, as lim x (x) = a + 3, () = lim x +(x) = + a + b it can be deduced that the unction will be continuous at that point when: a + 3 = + a + b a + b =. On the other hand, supposing the the unction is continuous at x =, it will have a derivative at that point i: a = () = +() = + a a =. So the unction will be continuous and derivable at x = when a = b =. b) By the mean value theorem we know that: There exists c (, 3) : (3) ( ) = (c)(3 ( )). Taking into account that a = b =, the ormer equation is equivalent to ( ) ( + 3) = (c). In other words: (c) =. When x, (x) =, so it is not possible that c. when x >, (x) = x + = x = 3. So the only possible value is c = 3.

6 (5) Given a continuous, concave unction : [0, 3] R and satisying: (0) =, () = 3, (3) = 0, () = 0. Then: (a) Draw approximately the set A = {(x, y) : 0 x 3, 0 y (x)} and ind the maximal and minimal elements, the maximum and the minimum o A i they exist. (b) Calculate the best lower (or by deault) approximation o the area o the given set. Hint or a: Pareto order is deined by: (x 0, y 0 ) P (x, y ) x 0 x, y 0 y. Hint or b: consider the segments that unite the points (0, ), (, 3) and (3, 0). 0.6 points part a); 0. points part b) a) As (x) is concave and () = 0, it means that the unction is increasing on the interval [0, ] and decreasing on the interval [, 3]. So, the drawing o A will be, approximately, this way: 3 (,3) (3,3) 3 by this graph, the Pareto order describes the set in the ollowing way: there is not a maximum o A, maximal points(a) = {(x, (x)) : x 3}. minimum(a) = minimal points(a) = {(0, 0)}. b) As the unction is concave, the segment h (x) that unites the points (0, ) and (, 3) remains below the graph o the unction, as the next igure shows: 3 (,3) (3,3) 3 So, as h (x) = + x, it can be deduced that: h (x)dx = 5 (x)dx 0 0 Analogously, the segment h (x) that unites the points (, 3) and (3, 0) also remains below the graph o the unction. So, as h (x) = ( 3 )(x 3), it can be deduced that: h (x)dx = 3 (x)dx So = Area (A)

7 (6) Given the unction h(x) = ln x, i x > 0, then: x3 (a) Find the primitive o h(x) whose value at x = is. (b) Let g be a continuous unction such that h(x) g(x) h(x) i x and let G(x) = g(t)dt. Find the best approximation o L, being y = L the horizontal asymptote o G(x). Hint or b: irst o all, prove the existence o L and compute its approximate value, supposing that lim h(t)dt = and proving / using the monotonicity o G(x). Secondly, check that lim h(t)dt = 0.5 points part a); 0.5 points part b) a) Let H(x) = ln x dx be the indeinite primitive o h(x). x3 Integrating by parts, and calling (x) = x 3, g(x) = ln x, you obtain: x 3 ln xdx = x ln x x x dx = x ln x + x 3 dx = = x ln x + (x ) + C = ( )x ( ln x + ) + C And now, as H () = + C = = C = 5. So we have H (x) = ( )x ( ln x + ) b) As h(t)dt g(t)dt h(t), and as lim h(t)dt =, it ollows that L = lim g(t)dt, i it exists, it will satisy L. So L = 3 8 will be the best approximation, with a maximum error o 8. On the other hand, as G(x) = g(t)dt is increasing, because its derivative, g(x), is positive, and also G(x) is upperly bounded, it can be deduced that L = To prove the hint, as H (x) = lim g(t)dt exist. h(t)dt is the primitive o h(x) that takes the value 0 in x =, then the mentioned primitive is exactly equal to the other primitive H(x) obtained in part a), except in one unit. So, H (x) = ( )x ( ln x + ) +. And now, lim h(t)dt = = lim H (x) = + ( ) lim = + ( ) lim /x x =. ( ln x + )/x =(applying the L Hopital s rule)= The ollowing graphs o g(x) and G(x) can help to understand the situation: y=g(x) y=h(x) / L / y=h (x) y=g(x) y=h(x) y=h (x)