MATH 211 Quiz: MATH 211 NIU s03Quiz7 8Solution
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Math 211 c (j. beachy: (6 pts) find f (x) for f (x) = x4ex2+1. f (x) = 4x3. Ex2+1(2x: (6 pts; p299 #63) find f (x) for f (x) = e. Optional simpli cation: f (x) = (4x3 + 2x5)ex2+1. Simplify rst: f (x) = e x + ex = ex1/2. + e(1/2)x f (x) = ex1/2 (1/2)x 1/2 + e(1/2)x(1/2) Ex: (8 pts) for the function f (x) = xe2x, nd the absolute maximum and absolute minimum values on the interval [ 1, 0]. We need to test the values of f (x) at the critical points and end points. To nd f (x), we need to use the product rule. f (x) = 1 e2x + x e2x(2) = e2x + 2xe2x = (1 + 2x)e2x (1 + 2x)e2x = 0. We can cancel the term e2x, since it is never zero. f ( 1) = ( 1)e2( 1) = .
