QUIZ ON CHAPTER 4 - SOLUTIONS APPLICATIONS OF DERIVATIVES; MATH 150 FALL 2016 KUNIYUKI 105 POINTS TOTAL, BUT 100 POINTS = 100%

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1 QUIZ ON CHAPTER - SOLUTIONS APPLICATIONS OF DERIVATIVES; MATH 150 FALL 016 KUNIYUKI 105 POINTS TOTAL, BUT 100 POINTS = 100% = x + 5 1) Consider f x and the grah of y = f x in the usual xy-lane in 16 x arts a) through f). If the answer to a art is none, write NONE. (5 oints) a) Find and box in all critical numbers of f. Show all work. (9 oints) f x = Dom( f ) = x x ± = { }, since those are the real values for which 16 x 0. Hi D( Lo) 16 x = D x ( x + 5) x + 5 D x ( 16 x ) ( Lo) ( 16 x ) x + 5 x ( 16 x ) ( x 10x) = = 16 x + x +10x ( 16 x ) ( 16 x ) ( 16 x ) Lo D Hi 16 x 1 = x +10x +16 = ( 16 x ) ( x + 8) x + ( 16 x ) The real values of x that make f DNE (namely, ± ) are not in Dom f cannot be critical numbers (CNs) of f. Now, solve f ( x) = 0 for real x in Dom f f ( x) = 0 ( x + 8) ( x + ) = 0 ( 16 x ) ( x + 8) x +. = 0, 16 x 0; that is, x ± x = 8 or x = Our two real solutions are in Dom f The critical numbers (CNs) of f are: 8 and. b) Find the x-intercet(s) on the grah of y = f x, so they, so they are critical numbers (CNs) of f.. ( oints) Find the real zeros of f. (We set y = 0, or f ( x) = 0, and solve for x.) x x = 0 The x-intercet is at ( 5, 0). x + 5 = 0 and 16 x 0 x = 5

2 c) Find the y-intercet on the grah of y = f ( x). ( oints) (We set x = 0 and solve for y. We evaluate:) f ( 0) = ( 0) = 5 5, so the y-intercet is at 0, d) Find the equation(s) of any horizontal asymtote(s) for the grah of y = f x. (3 oints) x + 5 is rational and bottom-heavy ; the degree of the denominator () is greater 16 x than that of the numerator (1). The horizontal asymtote is at y = 0, the x-axis. e) Find the equation(s) of any vertical asymtote(s) for the grah of y = f x. ( oints) x + 5 The denominator, 16 x, has only and as its real zeros. is simlified; 16 x the numerator and denominator have no common real zeros and no nontrivial factors in common. The vertical asymtotes are at x = and x =. f) Find the absolute maximum oint x, y and the absolute minimum oint ( x, y) on the grah of y = f ( x), if x is restricted to the interval 3, 0. Indicate which oint is the absolute maximum oint and which is the absolute minimum oint. (5 oints) f is continuous on excet at and. In articular, f is continuous on the closed, bounded interval 3, 0, so the Extreme Value Theorem (EVT) alies, and there exist absolute maxima and minima on 3, 0. Our candidates for x are [the critical number (CN) of f in 3, 0 x ] and the interval endoints, 3 and 0. Answers f x a = 3 f ( 3) = f ( ) = 1 = 0.5 (, 0.5) is the A.Min.Pt. on 3, 0. b = 0 f ( 0) = 5 16 = ( 0, 0.315) is the A.Max.Pt. on 3, 0.

3 ) State Rolle s Theorem, including the hyotheses and the conclusion. Write the conclusion using the algebraic notation we used in class; don t just refer to tangent lines or secant lines. (8 oints) If a function f is continuous on a closed, bounded interval a,b the oen interval ( a,b), and if f ( a) = f ( b), then c a,b is, the tangent line is horizontal there). (We assume a < b.) 3) Sketch the grah of y = f ( x), where f x xy-lane. (7 oints) [ ] and is differentiable on such that f ( c) = 0 (that = x 3 3x x +13, in the usual Find all critical numbers of f and label them CNs. Find all oints at critical numbers. Indicate these oints on your grah. Find all inflection oints (if any) and label them IPs. Indicate these oints on your grah. Classify all oints at critical numbers as local maximum oints, local minimum oints, or neither. Find the y-intercet. You do not have to find x-intercets. Have your grah show where f is increasing / decreasing and where the grah is concave u / concave down. Justify with work, as in class. You may round off any non-integers to five significant digits. Show all stes, as we have done in class. Ste 1: f is a olynomial function. Therefore, For this f : Dom f =. f is continuous on. The grah has no vertical asymtotes (VAs) and no holes. Furthermore, since f is nonconstant and nonlinear, the grah has no horizontal asymtotes (HAs) and no slant asymtotes (SAs). The y-intercet is at the oint 0,13 term from the given olynomial. f is neither even nor odd. Ste : Find f ( x) and critical numbers (CNs) of f. f ( x) = 3x 6x = 3 x x 8. This is because f ( 0) = 13, the constant = 3 x + ( x ) f is continuous on. In articular, it is never undefined ( DNE ). The real solutions of f The CNs are and. ( x) = 0 are given by x = and x =, which are in Dom f.

4 Ste 3: Do a sign diagram for f and classify the oints at the CNs. f is continuous on, so the First Derivative Test (1 st DT) should aly wherever we have CNs. Both f and f are continuous on, so we use just the CNs as fenceosts on the real number line where f could change sign. x Test x = 3 Test x = 0 Test x = 5 f sign (see below) f Classify Points at CNs (Use 1 st DT) L.Max. Pt. L.Min. Pt. Sub into f ( x) to get y (, f ( ) ) (, 1) = ( 3) ( x + ) ( x ) ( 3) = ( + ) ( ) ( ) = + = ( + ) ( + ) ( ) = = ( + ) ( + ) ( + ) = + f x f f 0 f 5 (, f ( ) ) (, 67) The multilicities of the zeros of f are both odd (1), so signs alternate in our windows. Furthermore, the grah of y = f Ste : Sketch a skeleton grah for y = f ( x) (otional). Horizontal tangent lines are indicated in brown. ( x) is an uward-oening arabola. Ste 5: Find f ( x) and ossible inflection numbers (PINs). = 3x 6x ( x) = 6x 6 = 6( x 1) f x f f is continuous on. In articular, it is never undefined ( DNE ). The real solution of f The PIN is 1. ( x) = 0 is given by x = 1, which is in Dom f.

5 Ste 6: Do a sign diagram for f and find inflection oints (IPs). f, f, and f are continuous on, so we use just the PINs as fenceosts on the real number line where f could change sign. x Test x = 0 1 Test x = f sign (see below) + f grah Inflection Points (IPs)? Sub into f ( x) to get y CD CU f x Yes, IP ( f cont. here, concavity changes) ( 1, f ( 1) ) ( 1, 13) = ( 6) ( x 1) = ( + ) ( ) = = ( + ) ( + ) = + f 0 f The multilicity of x = 1 as a zero of f is 1 (odd), so f changes signs. Also, the grah of y = f ( x) is a rising line with x-intercet at 1, 0 Ste 7: Sketch the grah of y = f ( x). (Remember our skeleton grah!). Zooming out (in the long run), the grah will resemble a rising snake. The leading term of x 3 3x x +13 is x 3, which has odd degree and a ositive leading coefficient.

6 ) You do not have to show work for these roblems. (9 oints total) a) The following is true of the olynomial function f : f ( 3) =, f ( 3) = 0, f 3 maximum oint for the grah of y = f ( x). Box in one: The oint 3, True = 5. True or False: The oint ( 3, ) must be a local False is a local minimum oint by the Second Derivative Test. f ( 3) = 0. is concave u at [actually, on a neighborhood of ] x = 3. The grah of y = f x b) The following is true of the olynomial function g : g( 6) =, g ( 6) = 3, g 6 local maximum oint for the grah of y = g( x). Box in one: True =. True or False: The oint ( 6, ) must be a 6 is not a critical number (CN) of g, since g 6 Local maxima must be at critical numbers. False is neither 0 nor undefined ( DNE ). c) The function h has the interval 6, 10 as its domain, and h is continuous on that interval. True or False: On the interval 6, 10, there is an absolute minimum oint for the grah of y = h x True in the usual xy-lane. Box in one: False This is true by the Extreme Value Theorem (EVT), because h is continuous on the closed, bounded interval 6, 10. 5) Let s( t) be the height in feet (at time t in seconds) of a article that is moving along a vertical line. If s ( ) = 3 ft = 5 ft, and sec s, what is the article sec doing at (really, on a neighborhood of) t = seconds? Box in one: (3 oints) a) The article is rising and is seeding u. b) The article is rising and is slowing down. c) The article is falling and is seeding u. d) The article is falling and is slowing down. = v( ) > 0, so the article s height is increasing, and the article is rising. = v( ) > 0 and s ( ) = a( ) < 0, the height is increasing at a s Because s decreasing rate, and the article is slowing down. (v is velocity; a is acceleration.)

7 6) We want to aroximate a root (or zero) of x + x 1 using Newton s Method with x 1 = as our seed (our first aroximation). (1 oints total) a) Find x, which is our second aroximation using Newton s Method. When rounding, use five significant digits; round off your answer to four decimal laces. Let f x x 1 = = x + x 1. Then, f ( x) = 8x x = x 1 f x 1 f x 1 = f ( ) f ( ) = ( ) + ( ) 1 8( ) 3 +1 = 9 63 = b) Find x 3, which is our third aroximation using Newton s Method. When rounding, use five significant digits; round off your answer to four decimal laces. x 3 = x f x f x f ( = f ( ) + ( ) 1 8( ) 3 +1 = Note 1: The real roots (zeros) of f are 1 and about The two other comlex roots are imaginary. Note : x , x , x , x , x The last two iterates are equal to four decimal laces, so we would take as our final aroximation. Note 3: We could have simlified the iteration formula (for n + ): x n+1 = x n x + x n n 1 8x 3 n +1 = x 8x 3 +1 n n 8x 3 n +1 x + x n n 1 8x 3 n +1 = 6x +1 n 8x 3 n +1 7) Prove that, among all rectangles with fixed erimeter, where > 0, the largest in area is a square. (Kee in your work; don t just ick a numerical value for.) Show all work, and verify that a square is, in fact, the rectangle with the absolute maximum area, as in class. Do not use any shortcut recalculus formulas or methods when methods from calculus can be used instead. (1 oints) Ste 1: Read the roblem.

8 Ste : Set u a diagram, a table, etc. Let l and w be the dimensions of a rectangle with erimeter. Ste 3: Write the rimary equation. Maximize A, the area of the rectangle. A = l w Ste : Write any relevant secondary (constraint) equations. Perimeter = l + w Ste 5: Exress A in terms of only one variable, say l. Solve the constraint equation = l + w for w in terms of l, say: l + w = w = l w = l ( Revised Secondary / Constraint equation ) Exress A in terms of l. Incororate the constraint into the rimary equation: A = l w A = l l A = l l Let f ( l) = l l. f is our objective function. Ste 6: Write the feasible domain of f. We require l 0. We also require w 0, but we will rewrite this in terms of l : w 0 l 0 ( From the Revised Constraint Equation ) l Dom f = 0, l Note: l = 0 and w = 0 may not make ractical sense, but they are ermissible, and it is convenient to have a closed interval as our feasible domain. Maximize A; that is, Maximize f Find critical numbers (CNs) of f in 0,. Remember that is a constant! f ( l) = l l f ( l ) = l never undefined ("DNE") on 0,.

9 Solve f l = 0 for l in 0, : l = 0 = l l = l = Now, 0,, so it is a critical number (CN). If l =, then, by the Revised Constraint Equation: w = l = = and thus l = w. The corresonding rectangle is a square. Verify that this CN leads to the absolute maximum of f on the feasible domain. Method 1: Extreme Value Theorem (EVT) Method f is continuous on the closed, bounded domain 0,, so the Extreme Value Theorem (EVT) alies. There exists an absolute maximum of f on 0,. Our candidates for l are, which is the critical number (CN) of f in 0,, and the endoints of the interval 0,. l a = 0 f 0 b = f ( l) Comments = 0 f = 16 f = 0 Method a: General Second Derivative aroach f ( l) = l f ( l) = f l < 0, l 0, There is an absolute maximum of f on 0, at., including l =.

10 The grah of y = f l is concave down on 0,, and f is continuous on 0,, so there must be an absolute maximum on 0, at l =, the sole critical number (CN) of f. Method b: Second Derivative Test aroach f f = 0, so we may aly the Second Derivative Test at l =. = < 0, so the grah of y = f l is concave down at (actually, on a neighborhood of ) l =, and there must be a local maximum there. Also, f is continuous on the feasible domain, 0,, and is the only critical number (CN) of f in 0,. Therefore, the local maximum of f at l = must also be an absolute maximum on 0,. Method 3: First Derivative aroach f f f ( l) = l ( l) > 0 on the l-interval 0,, ( l) < 0 on the l-interval f is continuous on 0,.,, and Therefore, there must be an absolute maximum on 0, at l =. l 0,, f sign + 0 f f is cont. on 0,. A.Max.