MATH 140 Lecture 26: Exam Review

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Review problems f ( x)=x3 3x +6for 0 x 3 . Show work. f " ( x)=3 x2 3=3(x2 1)=3( x+1) ( x 1) f " (x)=0at x= 1 o: using section 4. 1: f (0)=6,f (1)=4, f (3 )=24, thus, the minimum value is f (1)=4 and the maximum value is f (3)=24 . State carefully the mean value theorem: let f(x) be continuous on an interval [a, b] and differentiable on (a, b); then there is some value c in [a, b] such that f " (c)= f (b) f (a) b a. Leave your answer with any needed exponentials or logarithms: at t = 0, f(0) = 8 grams; f(4) = 6 grams f (t )=8ekt for k 0 o. Draw a o o o o y=9 x2 for 3 x 3. A=2xy a(x )=2x(9 x2)=18x 2 x3 0 x 3 maximize a.

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