Math Midterm Solutions

Transcription

1 Math 50 - Midterm Solutions November 4, 009. a) If f ) > 0 for all in a, b), then the graph of f is concave upward on a, b). If f ) < 0 for all in a, b), then the graph of f is downward on a, b). This is the concavity test on the page 9 of your tetbook. b) Differentiate implicitly and isolate y. + y = 4 + yy = 0 yy = 4 y = y If y is to be, then we must have y =. Let s plug this into the original equation. + ) = + 4 = 6 = = 6 = ± 6 The corresponding values of y are y = 6. satisfying the condition are 6, ), ), So the two points

2 c) We first make the substitution y = π. π π ) tan = y tany + π y 0 ) = y siny + π ) y 0 cosy + π ) Now we use the fact that cosy + π ) = sin y to get y 0 y sin y siny + π ) = siny + π y 0 ) = sinπ ) =. An alternative solution is to use L Hospital Rule.. a) If f has a local maimum or minimum at c, and if f c) eists, then f c) = 0. See Theorem 4 on the page 73 of the tetbook. b) f ) = 3 6 = 3 ) = 0 We get two critical points, = 0 and =. Since the domain of definition is a closed interval, we should also check the endpoints, = and = 4. So we plug these four values into the function and determine the ma and min. c) f ) = = 3 8 f0) = = f) = 8 + = 3 f4) = = 7 Hence the absolute maimum is 7 and occurs at = 4 while the absolute minimum is 3 and occurs at =. This problem is the Eample 8 on the page 75 of the tetbook. ) f ) = e arctan + ) ) f ) = e arctan + + e arctan + ) = = e arctan + ) f ) is defined for all and its only zero occurs at =. f is positive on, ) and negative on, ), hence f is, respectvely, concave up and concave down on these intervals.

3 3. a) If f ) = 0 for all in the interval a, b), then f is constant on a, b). This is Theorem 5 on the page 84 of the tetbook. b) We can factorize = ), so the equation has at least one solution, = 0. Now look at the function g) = It can be readily seen that g) is positive for 0 and negative for sufficiently large negative values take = for instance). Since g is continuous, the Intermediate Value theorem guarantees that it has at least one root, which has to be negative. In fact, it has precisely one root, because if it had two, say and, both of which would have to be negative, Rolle s theorem would guarantee the eistence of a number c in, ) such that g c) = 0. This is impossible, as g ) = , which is nonzero for < 0. We conclude that the original equation = 0 has two solutions. c) We write and take sin ) = e lnsin ) ) = e lnsin ), lnsin ) lnsin ) = /. The indeterminate form of this last it is [ / ], hence we can use L Hospital s rule. lnsin ) = 0 + / 0 + cos sin / = cos 0 + = cos ) sin = 0 = 0 Using the substitution law for its, we have ) = sin 0 +sin ) = 0 e lnsin ) = e 0+ lnsin ) = e 0 = If the it is to eist and equal 0, we must have sin) b ) = a. 3

4 We can write sin) b ) sin) + b = 0 3. This last it has indeterminate form [0/0], so we can apply L Hospital s rule. sin) + b cos) + b 0 3 = 0 3 If b is anything other than, putting = 0 in the numerator we get a nonzero value + b and the it is either or, and so a cannot be finite. Hence b must be. Putting b =, we again get the indeterminate form [0/0] and apply L Hospital s rule. cos) 4 sin) 0 3 = = sin) = It follows that a = 4 3, and so our answer is This problem was done in class. 5. Let f) = arcsin + at some arbitrary point, say =. a = 4 3, b =. ), g) = arctan π. Let s compute f and g f) = arcsin 0 = 0, g) = arctan π = π 4 ) π = 0 Now our strategy is to show that f ) g ) = 0 for all 0. Then, by 3.a), it will follow that f) g) = α for all > 0, where α is some constant, and by continuity this will also hold for = 0. Plugging in as above, we see that f) g) = 0, so that α is 0 and f) = g) for 0, which we want to show. f ) = = + + The result follows. ) ) ) = + This problem was done in class. + ) + ) ) + ) = + ) = + ) ) + = = 4 + = + ) = g ) 4

5 6. Let f) = +. f is continuous and differentiable on, ), with f ) =. Now fi > 0. By the Mean Value theorem, there eists + a number c in the interval 0, ) such that Now, +c <, because Finally, f f) f0) c) = c = + c + = + c > 0 + c > + c > + c <. + = + + c = + < + + c. 7. If we draw the graph of the curve y = and let, y) denote the coordinates of the upper right corner of the rectangle, it s very easy to see that the area of the rectangle is given by A = y. Because the upper right corner lies on the curve by assumption, the coordinates are actually, ), and so the area as a function of is given by A) = ) = By the conditions of the problem 0 and A) 0, and so the function we want to maimize is A) for the values of in the interval [0, 009) 00 ]. A ) = = = = = = 0 ) /00 This is the critical point of A). Note that the function f) = 00 is strictly increasing on 0, ), hence A ) = is strictly decreasing for > 0. So A ) changes from positive to negative at 5

6 = 009 /00, 0) and by the First Derivative test, this point is a local maimum. Since A0) = A009) 00 ) = 0, the point = 009 /00 0) is also the point of the absolute maimum of A) on the interval [0, 009) 00 ]. The dimensions of the rectangle are by , and our answer is ) / )