Math Practice Exam 3 - solutions

Transcription

1 Math Practice Exam 3 - solutions Problem 1 Consider the function h(x) = (9x 2 33x 25)e 3x+1. a) Find h (x). b) Find all values of x where h (x) is zero ( critical values ). c) Using the sign pattern of h (x), tell which of the x-vales in part b) is a local maximum and which a local minimum. Solution. First, for a) use the product rule and chain rule. h (x) = (18x 33)e 3x+1 + 3(9x 2 33x 25)e 3x+1 = 27(x 2 3x 4)e 3x+1. b) the exponential is never zero, so critical values are solutions of x 2 3x 4 = 0 namely x = 4 and x = 1. c) The sign pattern + + of h (x) reveals that x = 1 is a local maximum and x = 4 is a local minimum. This function is challenging to plot because it varies so much in scale! to really see what is going on near x = 1, we would need to zoom in to that region. Problem 2 a) Solve 7 exp( 0.4t + 3) 5 = 0

2 exactly and to 3 digits after the decimal point (see 5.3). b) Solve for x (see note that the logarithm is to base 10 here!) 5 log 10 (x + 2) = 15. Solution. a) Add 5 on both sides, divide by 7, take ln on both sides to get 0.4t + 3 = ln(5/7). Then solve for t, t = 3 ln(5/7) The first expression above is what we call an exact answer. The second is then the corresponding rounded decimal fraction. b) Divide by 5, use both sides as exponents of 10, so x + 2 = 10 log(x+2) = 10 3 and x = 998. Problem 3 Find the derivative of the following functions. Indicate whenever you are using one of the Big Three (Product Rule, Quotient Rule, Chain Rule) using the letters PR, QR, CR. Use only one of these per line (sum/difference and constant factor rules can go unmentioned). F (x) = e 5x cos(2x) G(x) = ln x 3x + 1 H(x) = e 5x3 2x Solution. For F (x), we need first the Product Rule and then the Chain Rule for each of the factors. With u(x) = e 5x and v(x) = cos 2x, we get u (x) = 5e 5x v (x) = 2 sin 2x F (x) = u v + uv PR! = 5e 5x cos(2x) 2e 5x sin(2x). For G(x), we need first the Quotient Rule and then only a couple of Rules

3 that we don t need to mention. With u(x) = ln x and v(x) = 3x + 1, we get u (x) = 1 x v (x) = 3 G (x) = u v uv = v 2 QR! (3x + 1)/x 3 ln x (3x + 1) 2. For H(x), we need first the Chain Rule (exponential Chain Rule here) and then only a couple of Rules that we don t need to mention. With u(x) = 5x 3 2x, we get u (x) = 15x 2 2 H (x) = e u(x) u (x) CR! = e 5x3 2x (15x 2 2). Problem 4 A function y = f(x) satisfies the equation x 3 + xy 2 = 5e y 2. It is also known that when x = 1, y = 2. a) Find a formula for the derivative of y = f(x) using implicit differentiation. b) Find f (1) using part a). c) Find an equation for the tangent line to the graph of y = f(x) at the point (1, 2) using part b). Solution. a) We differentiate both sides of the above equation, using the Product Rule for xy 2 and then the Chain Rule for the derivative of y 2 : (xy 2 ) = y 2 + x(2y)y. There is another Chain Rule needed to deal with e y 2. Altogether, we get 3x 2 + y 2 + 2xyy = 5e y 2 y. Now the goal is to solve for y. Subtract 3x 2 + y 2 from both sides to move it over to the right. Subtract 5e y 2 y from both sides to move it over to the left. Then factor out y on the left: y (2xy 5e y 2 ) = (3x 2 + y 2 ).

4 Finally, divide by the parenthesis on the left so we see y all by itself, y = 3x2 + y 2 2xy 5e y 2. b) Substitute x = 1 and y = 2 in the above formula for y. This gives y (1) = 7. c) We know the slope is 7, and it passes through the point (1, 2). So it has equation y = 7(x 1) + 2. Plot of the function from Problem 4, with tangent line at (1, 2). Problem 5 a) Find all solutions for x in [0, 4π] of 8 cos(x) = 0 Give exact answers as multiples of π, no rounded decimal fractions. b) Describe all solutions for x in [0, 10] of sin(3x + 1) = 0.2. Solution. a) Divide by 8, solve for cos x, so cos x = 2 2. Then x 1 = arccos( 2/2) = 3π/4 is a solution. Another solution is x 2 = 2π x 1 = 5π 4.

5 All solutions are then x 1 + 2πn, x 2 + 2πn for all integers n. b) We put t = 3x + 1 and first find possible values of t: t 1 = arcsin(0.2) 0.20 is the one solution in [ π/2, π/2], and t 2 = π t is another one which we get from the symmetry of the sine graph (you could make a quick sketch of a unit circle with the corresponding points on it, the sine function is the y-coordinate of these points). All possibilities for t are then t 1 + 2πn, t 2 + 2πn for integers n. Going back to we get the corresponding solutions x = t 1 3 x x Finally, since this function is periodic with period 2π/3, all solutions are of the form x = x 1 + 2πn/3 or x = x 2 + 2πn/3. Making a list of these values, we select the ones in [0, 10] as n x 1 + 2πn/3 N/A x 2 + 2πn/ In the plot below, solutions are the x-values where the red line crosses the sine graph.

6 Problem 6 Consider the function f(x) = 4 cos(3x + 5). a) Find the equation for the tangent line to the graph of f(x) at x = 2. Round all numbers to 2 digits after the decimal point. b) Use the equation found in a) to approximate f(1.97). Solution. a) First, the derivative of f(x) is f (x) = 12 sin(3x + 5). (use the chain rule!). For x = 2, we get f (2) = 12 sin and f(2) So the desired tangent line has equation b) Plug in x = 1.97 to get y = 12(x 2) y = 12(1.97 2) = Note that the true value is , we have a very accurate approximation.