FDWKCalcSM_ch4

FDWKCalcSM_ch4 - Section 4.1 161 81 Continued(e v(t = 3t 2...

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Section 4.1 161 81. Continued (e) vt t t t () =− > > > 32 0 6 2 2 2 82. (a) d dx ee du dx ux uu == where d dx ee ee xxxx + = −− 22 (b) d dx = + (c) y . 1 2 1 543 11 = + = = = + = y yx y . .( ). . 1 2 1 175 1 175 1 1 543 117 5 0 368 x + . (d) m m 2 1 1 175 0 851 . . − + + 0 851 1 1 543 0 851 2 394 .. (e) y xx 0 2 0 0 = = xxx or 83. (a) 10 2 −> x 2 1 >−<< , (b) = fx d dx xu x l n ( ) d dx u u du dx du dx x x x ln( ) = 1 2 2 1 2 (c) 1 1 2 −>− << , (d) = y 1 2 2 1 2 1 1 2 1 34 43 2 / / Chapter 4 Applications of Derivatives Section 4.1 Extreme Values of Functions (pp. 187–195) Exploration 1 Finding Extreme Values 1. From the graph we can see that there are three critical points: x 101 ,,. Critical point values: ff f () . ,( ) ,( ) . −= = = 5 0010 5 Endpoint values: = 20 420 4 Thus f has absolute maximum value of 0.5 at x = –1 and x = 1, absolute minimum value of 0 at x = 0, and local minimum value of 0.4 at x = –2 and x = 2. 2. The graph of f has zeros at x = –1 and x = 1 where the graph of f has local extreme values. The graph of f is not defined at x = 0, another extreme value of the graph of f . 3. Using the chain rule and d dx x x x = , we find df dx x x x x = + i 1 1 2 . Quick Review 4.1 1. = x d dx x x ( ) 1 24 4 1 i 2. = d dx d dx x ( ) // 29 9 9 21 2 23 2 2 i =− − = ( ) / / 92 2 9 2 232 x x 3. gx x d dx x x x s i n( l n) l n sin (ln ) i 4. hx e d dx xe i 5. Graph (c), since this is the only graph that has positive slope at c. 6. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c. 7. Graph (d), since this is the only graph representing a function that is differentiable at b but not at a. 8. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b . 9. As →− −+ 39 0 2 ,. Therefore, lim ( ) . x =∞ 3 10. ++ 0 2 Therefore, lim ( ) . x →− + 3

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162 Section 4.1 11. (a) d dx xxx () 32 232 −=− =− = f 13 1 21 2 (b) d dx x += 21 = f 31 (c) Left-hand derivative: lim ( ) [( ) ( )] hh fh f h +− = +−+− 00 3 22 2 2 2 4 61 0 0 10 0 0 2 h h h h h = ++ =+ + = lim ( ) Right-hand derivative: ( ) ) ] f h h h + + = ++− = 2 2 4 h h h h + + = = 0 0 1 1 Since the left-and right-hand derivatives are not equal, f 2 is underfined. 12. (a) The domain is x 2. (See the solution for 11.(c)). (b) = −< > fx xx x , , 2 12 2 Section 4.1 Exercises 1. Minima at (–2, 0) and (2, 0), maximum at (0, 2) 2. Local minimum at (–1, 0), local maximum at (1, 0) 3. Maximum at (0, 5) Note that there is no minimum since the endpoint (2, 0) is excluded from the graph. 4. Local maximum at (–3, 0), local minimum at (2, 0), maximum at (1, 2), minimum at (0, –1) 5. Maximum at x = b , minimum at xc = 2 ; The Extreme Value Theorem applies because f is continuous on [ a , b ], so both the maximum and minimum exist. 6. Maximum at x = c , minimum at x = b ; The Extreme Value Theorem applies because f is continuous on [ a , b ], so both the maximum and minimum exist.
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FDWKCalcSM_ch4 - Section 4.1 161 81 Continued(e v(t = 3t 2...

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