Loyola University Chicago Math 131, Section 009, Fall 2008 Midterm 2 Name (print): Signature: Please do not start working until instructed to do so. You have 50 minutes. You must show your work to receive full credit. Calculators are OK. You may use one double-sided 8.5 by 11 sheet of handwritten (by you) notes. Problem 1. Problem 2. Problem 3. Problem 4. Problem 5. Total.
Problem 1.(10 points total) Find derivatives of the following functions. Give your answer in exact form (do not use decimals). ( ) a.(5 points) f(x) = arctan 3x 2 + x Solution: use chain rule ( ( )) f (x) = 2 arctan 3x 2 1 + x = 1 + (3x 2 2 (6x + 1) + x) b.(5 points) f(x) = ( ) 4x 1 3 sin(x) Solution: use chain rule and quotient rule f (x) = ( (4x ) ) 1 3 ( ) 4x 1 2 4 sin(x) (4x 1) cos(x) = 3 sin(x) sin(x) sin 2 (x)
Problem 2.(10 points) possible value of x + y? The product of two positive numbers x and y is 270. What is the smallest We need to minimize x + y knowing that x, y are positive and xy = 270. So y = 270/x and we need to minimize f(x) = x + 270/x with 0 < x < 270. So f (x) = 1 270 x 2 and f (x) = 0 when x = 270. This is the only critical point. To make sure it is a minimum, try f (x) = 540 x 3 and note that f ( 270) > 0. So x = 270, y = 270/ 270 = 270 and the smallest possible value of x + y is 2 270.
Problem 3.(10 points total) Find two points (x, y) on the circle x 2 + y 2 = 1 at which the tangent line to the circle has slope 2. Solution: Implicit differentiation gives 2x + 2y dy dx = 0, dy dx = x y. We need dy dx = 2, so x y = 2, so x = 2y. Plug this in to the equation of the circle: (2y) 2 + y 2 = 1, 4y 2 + y 2 = 1, 5y 2 = 1, y 2 = 1 5. So, y = 1/ 5 or 1/ 5. The two points on the circle are then ( ) 2 1 5,, 5 ( 2 5, 1 5 )
Problem 4.(10 points total) The derivative of a certain function f(x) is given by the formula f (x) = x 2 (x 2)(x + 3). a.(6 points) Find all critical points of f(x) and clearly identify them as a local minimum, local maximum, or neither. Justify your answers. Solution: f (x) = 0 if x = 0 or x = 2 or x = 3. These are the critical points. To determine if they are local minima, maxima, or neither, check the sign of f (x) in between the critical points. f ( 4) = 16( 6)( 1) > 0, f ( 1) = 1( 3)(2) < 0, f (1) = 1( 1)(4) < 0, f (3) = 9(1)(6) > 0. Thus, x = 3 is a local maximum, x = 0 is not a local maximum nor a local minimum, x = 2 is a local minimum. b.(4 points) of f(x). Find the second derivative f (x) of the function f(x), and at least one inflection point Solution: First, simplify f (x) = x 2 (x 2 + x 6) = x 4 + x 3 6x 2. Now, find f (x): f (x) = 4x 3 + 3x 2 12x = x(4x 2 + 3x 12). So, f (x) = 0 when x = 0. This is a good candidate for an inflection point. To see that it really is an inflection point, note that 4x 2 + 3x 12 is negative when x is close to 0, but x changes sign from negative to positive at x = 0. So, f (x) changes sign from positive to negative at x = 0.
Problem 5.(10 points) Waterpark management decided to install a rectangular billboard with a picture of a dolphin under a popular waterslide. The end of the slide is three meters away from the wall where it starts (see picture). According to the manual, the slide has the shape of a parabola with equation y = 2x 2. Compute the area of the largest possible billboard that can fit into the space between the slide and the wall. 16 12 8 4 1 2 3 Solution: Let x be the x-coordinate of the lower left-hand corner of the billboard. Then the horizontal side of the billboard is 3 x, while the vertical side of the billboard is 2x 2. So, the area of the billboard is A(x) = (3 x)2x 2 = 6x 2 2x 3. To maximize this function on the interval [0, 3], find A (x) = 12x 6x 2 = 6x(2 x). So A (x) = 0 if x = 0 or x = 2. x = 0 gives zero area, so it is obviously not the maximum. The other endpoint, x = 3, also gives zero area. A(2) = 8, so this is the global maximum. (One could also check A (x) = 12 12x at x = 2. One gets A (2) = 12 < 0 so x = 2 is a local maximum.) So, the maximum area of the billboard is 8.