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Chapter 4 5 - 31 32 33 34 35 36 37 38 39 41 42 SECTION 4.1...

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Unformatted text preview: 31. 32. 33. 34. 35. 36. 37. 38. 39. 41. 42. SECTION 4.1 MAXIMUM AND MINIMUM VALUES 267 f($) = 5332 + 4m :> f’(m)210;r + 4. f’(z) 2 0 2 a: 2 β€”Β§. so β€”% is the only critical number. ftcc)=w3+ac2β€”w => f/(x)23x2+2mβ€”1. f’(e)=0 2 (m+1)(3mβ€”1)20 => m=β€”1.Β§. These are the only critical numbers. f(m) = e3 + 39:2 β€” 24:1: 2 f/(at) = 33:2 + 62: β€” 24 2 3(332 + 2x β€” 8). f'(:c) 2 0 2> 3(m + 4)(:c β€” 2) 2 0 2 a: 2 41. 2. These are the only critical numbers. m 2 #. Neither of these is a real number. Thus. there are no critical numbers. 803) = 35* +4253 β€” 6t2 2 s’(t) : 121:3 +12r2 212:. 5'(t) : 0 2 1225(232 + t β€” 1) 2 : Oor t2 + t β€” 1 2 0. Using the quadratic formula to solve the latter equation gives us a 2 _ 4 _1 a 21:: 5 t : 1i 1 (1X ) : 1 i β€œ5 2 0.618. β€”1.618. The three critical numbers are 0, 4. 2(1) 2 2 2 2 z+1 z +Z+1)12(z+1)(2z+1) 22 -22 f(Z):2β€”β€” β€œ5 f’(Z)β€”( 2 2 β€” 2 220 β€˜i’ Z+Z+1 (z +z+1) (z +z+1) 2(2 + 2) 2 0 2> z 2 0. β€”2 are the critical numbers. (Note that 22 + z + 1 2 0 since the discriminant < 0.) 2mβ€”lβ€”3 trapβ€”320 2 rfar>β€”g 9(w)=12$+3|: . => ’(ILβ€˜ : . 3 ~(2m+3)1f2mβ€”β€”3<0 221fcc<β€”Β§ g’(ac) is never 0. but g’(:c) does not exist for :1: 2 β€”%. so ~g is the only critical number. ()_ 1/3_ β€”2/3 :> /()_1 β€”2/3+g β€”5/3_l β€”5/3( +2)_$+2 gac2a: x 932233: 32: 23:0 :1: 23%,)”. g’(~2) 2 0 and g’(0) does not exist, but 0 is not in the domain of 9, so the only critical number is 22. g(t) 2 5152/3 + t5/3 2 g’(t) 2 13913β€”1/3 + 319/3. g/(O) does not exist. sot 2 0 is a critical number. g’(t) 2 Β§t_1/3(2 + t) 2 0 42 t 2 β€”2. sot 2 ~2 is also a critical number. 1 g(t) 2 x/{(1 β€” t) 2 151/2 β€” 753/2 2 g’(t) 2 m β€” gfi. g’(0) does not exist. sot 2 Dis a critical number. , 1 β€” 3t 1 1 . . . 0 β€” g (t) 2 2 fl t 2 5. sot 2 3 is also a critical number. F($) 2$4/5(xβ€”4)2 => F'(m)2x4/5-2(xe4)+(1’—4)2-Emil/52Γ©wβ€˜l/5(x~4)[5-m-2+($β€”4)-4] m~414mβ€”16 2mβ€”4 7xβ€”8 : % 2 % 2 OWhenm2 4. %;and F'(0) does not exist. Critical numbers are 0. g. 4. C(m) = V3 3:2 β€” m 2 G'(:c) 2 %(IL'2 ~ are) 32” (2w ~1). G'(a:) does not exist when $2 β€” :2: 2 0. that is. when w 2 0 or 1. G’(m) 2 0 <2 23: β€” 1 2 0 <2 w 2 %. So the critical numbers are x 2 0.1.1. to ...
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