1. Let f(x) = 3 + x + 5. Math 105: Review fr Exam I - Slutins (a) What is the natural dmain f f? [ 5, ), which means all reals greater than r equal t 5 (b) What is the range f f? [3, ), which means all reals greater than r equal t 3 2. Fr the graph f f shwn, answer the fllwing. (a) Evaluate the fllwing. f(x) i. f ( 2) = 0 ii. f(3) = 2 iii. lim f(x) = 2 x 3 iv. lim f(x) = 3 x 3 + v. lim f(x) des nt exist x 3 vi. lim f(x) = 0 x 2 (b) Where is f discntinuus? at x = 3 4 2 4 3 2 2 2 3 4 (c) Where des f fail t exist? at x = 3, 1, 0, 3 3. Let f(x) = 3x 2 2x. (a) Cmpute the average rate f change f f n the interval [2,2.1]. f(2.1) f(2) 2.1 2 = 9.03 8 0.1 = 10.3 (b) Using the limit definitin f the derivative, find f (x). f (x) h 0 f(x + h) f(x) h 3(x + h) 2 2(x + h) (3x 2 2x) 3x 2 + 6xh + 3h 2 2x 2h 3x 2 + 2x h(6x + 3h 2) (6x + 3h 2) h 0 = 6x 2 prvided this limit exists (c) Find the equatin f the tangent line t f at x = 2. We want y = mx + b. m = f (2) = 6 2 2 = 10, s y = 10x + b. When x = 2, y = f(2) = 3 2 2 2 2 = 8. Thus, 8 = 10 2 + b, s b = 12 and we have y = 10x 12. (d) Hw wuld the derivative f g(x) = f(x) + 5 cmpare t f (x)? The graph f y = f(x) + 5 is the graph f y = f(x) shifted vertically by 5 units, but this has n effect n the slpe f the graph, s g (x) = f (x).
(e) Hw wuld the derivative f h(x) = 5f(x) cmpare t f (x)? The graph f y = 5f(x) is the graph f y = f(x) stretched vertically by a factr f 5; this als results in slpes that are 5 times greater at any given x-value. Thus, h (x) = 5f (x). Nte that we get the same result by cnsidering ur derviative rule d dx [kf(x)] = kf (x) where k = 5. 4. Fill in the table shwing the graphical relatinships between f, f, and f. f psitive negative increasing decreasing cncave up cncave dwn f X X psitive negative increasing decreasing f X X X X psitive negative f 5. Given the graph f f, sketch a graph f f and a graph f F, an antiderivative f f such that F(0) = 2. f F(x) -2 f(x) f (x) Nte: The cncave up prtin in the middle f the graph f f is a perfect parabla, s its derivative (f ) is linear; since yu dn t knw the equatin fr f, yur graph f f may be cncave up/dwn there.
6. Shwn belw is a graph f f n its entire dmain. The graph is NOT f. At which x-value(s) (a) des f have a statinary pint? c, e, j (b) des f have a lcal max? e (c) des f have a lcal min? c (d) des f have a statinary pint? d, h, j (e) des f have a lcal max? d, j (f) des f have a lcal min? h (g) is f greatest? a (h) is f least? k (i) is f greatest? d (j) is f least? a (k) is f greatest? b (l) is f least? f (b) f decreasing? [a, c) (e, k] (c) f increasing? [a, d) (h, j) (d) f decreasing? (d, h) (j, k] (e) f cncave up? [a, d) (h, j) (f) f cncave dwn? (d, h) (j, k] a b c d f (x) e f g h i j k On what interval(s) is (a) f increasing? (c, e) 7. Suppse that T(t) gives the temperature in Lewistn as a functin f time. In each f the fllwing situatins, determine if the signs f T, T, and T are psitive, negative, zer, r unknwn. (a) The temperature is 60 degrees and falling steadily. The temperature is 60, s we knw T is psitive. The temperature is falling, s we knw T is negative. The temperature is falling steadily, s we knw the graph is linear, and T is zer. (b) The temperature is rising mre and mre slwly. We dn t knw whether the temperature is abve r belw zer, s the sign f T is unknwn. The temperature is rising, s we knw T is psitive. The temperature is rising mre and mre slwly, s we knw the graph f T is cncave dwn, and T is negative. (c) The temperature is 5 degrees and rising. The temperature is 5, s we knw T is negative. The temperature is rising, s we knw T is psitive. We dn t knw the cncavity f the graph f T, s the sign f T is unknwn. 8. The table belw gives sme values fr a functin f(x) whse derivative exists at all x. (a) Estimate f (1.05). f(1.1) f(1.0) 8.2 7.3 = 1.1 1.0 1.1 1.0 = 9 x 0.8 0.9 1.0 1.1 1.2 f(x) 5.0 6.2 7.3 8.2 9.0
(b) Based n the data, is f (1.0) psitive r negative? Using the same prcedure as in the previus part, we can make the fllwing estimates. f (0.85) 12 f (0.95) 11 f (1.05) 9 f (1.15) 8 We see frm these estimates that f (x) appears t be decreasing near x = 1. If f (x) is decreasing, then f (x) is negative (that is, f(x) is cncave dwn). 9. Find the derivatives f the fllwing. (a) y = 2 + 3x + x 4 + 5x 6 y = 3 + 4x 3 + 30x 5 (b) y = 6 x + 1 x + x 6 6 + 6 x + π 6 + 61/2 + 6x 1/6 First, rewrite y t make it easier t apply ur derivative rules: y = x 1/6 + x 6 + 1 6 x + 6x 1 + π 6 + 61/2 + 6 1/2 x 1/12 Nte that abve 6x 1/6 = 6 x 1/6 = 6 1/2 (x 1/6 ) 1/2 = 6 1/2 x 1/12. y = 1 6 x 5/6 + ( 6)x 7 + 1 6 + (6)( 1)(x 2 ) + 0 + 0 + 6 1/2 1 12 x 11/12 T clean this up, we use expnent rules: x a = 1 x a and xa/b = b a y = 1 6 6 x 5 6 x 7 + 1 6 6 x 2 + 6 12 12 x 11 10. Find antiderivatives f the fllwing. (a) y = π + 3x 2 antiderivative = πx + x 3 + C (b) y = 4x 5 1 x 6= 4x5 x 6 antiderivative = 4x6 6 x 5 5 + C = 2x6 3 + 1 5x 5 + C 11. Is y = 5x 3 a slutin t the differential equatin xy 3y = 0? The questin asks whether, when we plug in y and y, xy 3y will equal 0. We are given y = 5x 3, s y = 15x 2. xy 3y? = 0 x 15x 2 3 5x 3? = 0 15x 3 15x 3? = 0 0 = 0 This is true. Substitute y and y in the apprpriate places. S, y = 5x 3 is a slutin t the given differential equatin. 12. Slve the IVP (initial value prblem) 1 = x 3 y (x) if y(2) = 13. We begin by islating y (x). This gives y (x) = 1 + x 3 Next we find the antiderivative f y (x): y(x) = x + x4 4 + C.
Nw we plug in the 2 and the 13 t find the value f C. 13 = 2 + 24 4 + C 13 = 2 + 4 + C 11 = C S, the slutin t this IVP is y(x) = x + x4 4 + 11. See ld exams and quizzes at http://abacus.bates.edu/ etwne/mathresurces.html