Unformatted text preview: SECTION 4.1 MAXIMUM AND MINIMUM VALUES 273 73. (a) 11(r) = k(r0 — r)r2 : kr0r2 — kr3 => 1/(r) = 2kr0r — 3kr2. v’(r) : 0 => kr(2ro ~ 37") = 0
=> 7" = 0 or gm (but 0 is not in the interval). Evaluating v at éro. ﬁre. and m. we get 0671;) : ékrg.
v(§r0) = ékrg. and v(r0) : 0. Since 217 > §. 12 attains its maximum value at 7" : gm. This supports the statement in the text. (b) From part (a). the maximum value Of’U is 2%]97‘8. 74. 9(a) : 2 + (an i 5)3 => g’(:n) = 3(a: — 5)2 => g’(5) = 0, so 5 is a critical number. But 9(5) : 2 and 9
takes on values > 2 and values < 2 in any open interval containing 5. so 9 does not have a local maximum or minimum at 5. 75. f(9:) : $0101 —I— 3:51 + :2: +1 :> f’(m) : 101$100 + 513050 +1 2 1 for all m. so f’(:c) : 0 has no solution. Thus. ﬂat) has no critical number, so ﬂan) can have no local maximum or minimum. 76. Suppose that f has a minimum value at c. so f(ac) 2 f(c) for all 5c near 0. Then 9(a) : if(m) g —f(c) : g(c) for all a: near c. so 9(a) has a maximum value at c. 77. If f has a local minimum at c. then g(m) I —f(a:) has a local maximum at c. so g'(c) = 0 by the case of Fermat‘s Theorem proved in the text. Thus. f'(c) : —g’(c) : 0. 78. (a) f(:c) 2 ax3 + 172:2 + etc + d, a 7S 0. So f’(m) : 3am2 + 21353 —I— c is a quadratic and hence has either 2. 1. or 0
real roots. so f (3:) has either 2. 1 or 0 critical numbers. Case (1') (2 critical numbers): Case (ii) (1 critical number): Case (iii) (no critical number):
f(9c) : $3 2... 2 Has) : x3 => are) :223 +3.. :»
f'(a:) 2331c2 —3.som= —1.1 f’(;r) :3m2.som:0 f’(a:) :3x2—I—3.
are critical numbers. is the only critical number. so there are no real roots.
y (b) Since there are at most two critical numbers. it can have at most two local extreme values and by (i) this can occur. By (iii) it can have no local extreme value. However. if there is only one critical number. then there is no
local extreme value. ...
View
Full Document
 Spring '10
 Ban
 Calculus

Click to edit the document details