# MATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4.

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1 MATH 55 SOLUTIONS TO PRACTICE FINAL EXAM x 2 4.Compute x 2 x 2 5x + 6. When x 2, So x 2 4 x 2 5x + 6 = (x 2)(x + 2) (x 2)(x 3) = x + 2 x 3. x 2 4 x 2 x 2 5x + 6 = = 4. x Compute x + sin x. If x is close to but larger than, then the denominator sin x is a small positive number and x 2 9 is close to 9. So the quotient x2 9 sin x is a large negative number. So x 2 9 x + sin x =.

2 2 SOLUTIONS TO PRACTICE FINAL EXAM 3. Evaluate ( x 2 x x 2 + 5x). x 2 x x 2 + 5x) = ( x 2 x x2 x + x x 2 + 5x) + 5x x2 x + x 2 + 5x = (x 2 x) (x 2 + 5x) x2 x + x 2 + 5x = = 6x x 2 ( x ) + x 2 ( + 5 x ) 6x x x + x + 5 x 6 = x x 6 = = { ax + x <, Let f(x) = x 2 + x. For what constant a is f differentiable everywhere? f is clearly differentiable for x < and for x >. For x <, f (x) = a, so f (x) = a. For x >, f (x) = 2x, so f (x) =. For f to x x + be differentiable at, we need a = f (x) = f (x) =. x x +

3 tan 2x 5. Compute x sin 3x. SOLUTIONS TO PRACTICE FINAL EXAM 3 tan 2x x sin 3x = x 6. Compute x. sin 2x ( cos 2x sin 3x sin 2x = x x 2 sin 2x = x 2x cos 2x x sin 2x = 2 x 2x x = 2() 3 () = x2 + x +. 3x ) x sin 3x cos 2x 3x x 3 sin 3x cos 2x 3 3x x sin 3x Note x = x 2 for x <. Multiply the top and bottom of 4x 2 +x+ 3x by to get x 4x2 + x + = 4x x 2 + x (/x) + (/x2 ) =. 3x (3x ) 3 (/x) x 4 Hence the it as x is 3 = Compute the tangent line to the ellipse given by the equation x 2 + 4y 2 = 5 at the point (, ). If we take the derivative with respect to both sides of the equation we see 2x + 8y dy dy =, or = x. So the slope at (, ) is dx dx 4y. Thus the tangent line at (, ) is 4 or y + = (x ), 4 y = 4 x Let F (x) = f(g(x)). Compute F (2) using the following information: f( ) = 3, f(2) = 2, g( ) = 7, g(2) =, f ( ) = 2, f (2) = 8, g ( ) =, g (2) = 5.

4 4 SOLUTIONS TO PRACTICE FINAL EXAM. Using the chain rule, F (x) = f (g(x))g (x), so F (2) = f (g(2))g (2) = f ( ) 5 = 2 5 = 9. For y = (sin 4x) 8, compute y.. Using the chain rule a total of three times, we get y =8 (sin 4x) 7 d dx (sin(4x)) = 8 (sin d 4x)7 cos 4x dx 4x =32(sin 4x) 7 cos 4x.. How many inflection points does the curve y = x5 + x4 have? 5 4. First we note y = x 4 +x 3 and y = 4x 3 +3x 2 = x 2 (4x+3). Hence y = at x = and x = 3/4. However, y < in (, 3/4) and y > in ( 3/4, ) and (, ). Hence only 3/4 is the inflection point.. Compute the derivative y for the curve x 2 + y 2 = 2 + y at the point x = 4, y = 3.. Taking the derivative of both sides of the equation and using the chain rule gives /2(x 2 +y 2 ) /2 (2x+2yy ) = y. So evaluating at the point x = 4, y = 3, we get /2( ) /2 ( y ) = y gives /(8 + 6y ) = y which we can solve for to get y = A kite ft above the ground is flying horizontally (away from its holder) with a speed of 6ft/sec. At what rate is the angle between the string and the horizontal direction changing, when 2 ft of the string have been let out?. The kite is at a constant height of ft with a length of x ft away. So using some trigonometry, we see that if θ is the angle between the string and the horizontal direction, tan θ = /x. Taking the derivative with respect to t gives sec 2 θ dθ 2 dx = x = 6x 2 dt dt since the kite is flying away at 6ft/sec. When 2ft have been let out, sin θ = /2 and θ = π/6. At this value of θ we have 4/3 dθ = 6 dt or dθ dt = 2 3 = radians 25 second. 3. Find the linearization of f(x) = x 2 at a =.. The linearization of f(x) at a = is L(x) = f(a) + f (a)(x a). At a =, f(a) = 3 and f (x) = /2( x 2 ) /2 ( 2x) so f (a) = /3 and substituting back in gives f(x) = 3 + /3(x + ).

5 SOLUTIONS TO PRACTICE FINAL EXAM 5 4. Find all local maxima and minima of the function f(x) = 2 x x 2.. Taking the derivative of 2x x 2 for x > gives f = 2 2x which means a critical point is at x =. Taking the derivative again and using the second derivative test gives x = is a local maxima. Taking the derivative of 2x x 2 for x < gives f = 2 2x which means a critical point is at x =. Taking the derivative again and using the second derivative test gives x = is a local maxima. When x < the function is increasing and when x > the function increases for small values away from so it is easy to see that x = is a local minima. 5. Find all asymptotes of the curve y = 2x2 +x+ x.. The curve y = 2x2 +x+ is undefined at x = and so has a x vertical asymptote there. Now, 2x 2 + x + = ±, x ± x so there is no horizontal asymptote. Let s do long division to see if there is a slant asymptope. One gets 2x 2 + x + x = 2x x. Hence y = 2x+3 is a slant asymptope because as x ±, 2x+3+ 4 x goes to 2x Find all the points on the hyperbola y 2 x 2 = 4 that are closest to the point (2, ).. The distance formula between a point (x, y) and (2, ) is given by d(x, y) = y 2 + (x 2) 2. We want to consider only points (x, y) such that y 2 x 2 = 4. Solving for y 2, y 2 = x Substituting into the distance formula, we have d(x) = x (x 2) 2 = 2x 2 4x + 8. To minimize distance, we use the first derivative to find critical points. d 4x 4 (x) = 2 2x 2 4x + 8. If d (x) = then 4x 4 = and so x =. Note d (x) < for x < and d (x) > for x >. Hence d(x) decreases for x < and increases

6 6 SOLUTIONS TO PRACTICE FINAL EXAM for x >, and therefore d(x) realizes its global minimum at x =. Because a hyperbola is not an actual function, there can be more than one y associated with a particular x. Earlier we found y 2 = x Then y = x When x =, y = ± 5. So the two distance minimizing points on the hyperbola are (, ± 5). 7. A page of a book is to have a total area of 5 square inches, with inch margins at the top and sides, and a 2 inch margin at the bottom. Find the dimensions in inches of the page which will have the largest print area.. You should draw a picture for this problem. If l is the total length of the page, w is the total width, and A is the print area, then lw = 5, A = (l 2)(w 3). We want to maximize A so we want to substitute in for one of the variables so that we can take the derivative. Note l = 5/w. Therefore, and A(w) = ( 5 w 2)(w 3) = 5 45 w 2w + 6, A (w) = 45 w 2 2. If A (w) = then w 2 = 225 so w = 5. The first derivative test easily shows this gives a maximum area. Since l = 5/w, l =. 8. Newton s method is to be used to find a root of the equation x 3 x =. If x =, find x 2.. If we let f(x) = x 3 x then f (x) = 3x 2. Hence x 2 = x f(x ) f (x ) = f() f () = 2 = Express the it below as a definite integral. n ( ) π iπ n 4n sec2 4n i=. It is probably easiest to guess first that the sum is a straightforward Riemann Sum and only look for tricks if necessary. Normally in a Riemann sum to n, n is the number of divisions of the

7 SOLUTIONS TO PRACTICE FINAL EXAM 7 interval. We should also expect the π to be the length of the subintervals. Hence the total interval should be of length π. Then it is clear 4n 4 that iπ is simply the right-hand endpoint of the i-th interval. Hence 4n our it is simply π/4 sec 2 (x)dx. 2. If f(x) = 5x cos(u 2 )du, find f (x). We want to apply the Fundamental Theorem of Calculus, but cannot immediately. So we define g(x) = x cos(u 2 )du to get f(x) = g(5x). By the chain rule, f (x) = 5g (x). Now we can apply the Fundamental Theorem to g: f (x) = 5g (x) = 5 cos((5x) 2 ) = 5 cos(25x 2 ). 2. Evaluate the integral. π π x sin(x 2 )dx. Let u = x 2. Then du = 2xdx and x sin(x 2 )dx = 2 π sin(u)du = 2 cos(u) π = /2 + /2 =. 22. Which of the following integrals give the area of the region below the curve y = 2x and above the curve y = x 2 4x?. You should draw a picture. We find the intersection points of the two curves by solving 2x = x 2 4x x 2 6x = x(x 6) = x =, 6. For small x, x 2 4x <. In [, 6], the curve y = x 2 4x is below y = 2x. This can be seen by taking, say, x =. Therefore the area between the curves is given by 6 ( ) 2x (x 2 4x) dx. 23. Consider the area in the xy plane bounded by the curves y = and y = x x 2. If we rotate this area about x = 7, what integral gives the volume?. Draw a graph. Use the shell method. The two curves intersect when = x x 2. This happens for x =,. Since we are

8 8 SOLUTIONS TO PRACTICE FINAL EXAM rotating about the line x = 7 the radius of each shell will be 7 x. The height of the shell will be given by x x 2. Therefore the integral is 2π(7 x)(x x 2 ) dx = 2π (7 x)(x x 2 ) dx. 24. The plane region bounded by the curves y = 2 and y = 2+2x x 2 is rotated about the x axis. What integral gives the volume?. Draw a picture. Use the disk method. The two curves intersect when 2 = 2 + 2x x 2. Then x 2 2x =, x(x 2) =, and hence x =, 2. By testing a value, one sees that y = 2 + 2x x 2 is above y = 2 for x between and 2. Since we are rotating about the x-axis, the inner radius is 2 and the outer radius is 2 + 2x x 2. Hence the correct integral is 2 ) 2 ( ) π ((2 + 2x x 2 ) dx = π (2 + 2x x 2 ) 2 4 dx. 25. The function f(x) = 6 2x is continuous on the interval [, 8]. What is its average value on this interval?. By definition the average value is x dx. Toward finding the antiderivative of the integrand, we make the substitution u = 6 2x. du = 2. 6 u 2 2x dx = du = u3/2 = 3 (6 2x)3/2. Going back to our expression for average value, we have 8 6 2x dx = (6 2x)3/2 8 = 24 ( 63/2 ) = = 8 3.

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