sample-midterm2-with-solutions-v2

That f c 3 0 3 0 since f x 3x 2 6 x 1 we have

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Unformatted text preview: x) = x 3 - 3x 2 + x on [0,3]. Solution Since the two hypotheses of the Mean Value Theorem are satisfied by f ( x) = x3 - 3x 2 + x on [0,3], there must exist a number c in the open interval (0,3) such f (3) - f (0) 3 - 0 = = 1. that f '(c) = 3-0 3-0 Since f '( x) = 3x 2 - 6 x + 1, we have f '(c) = 3c 2 - 6c + 1 = 1 Þ 3c 2 - 6c = 0 Þ 3c(c - 2) = 0 Þ c = 0 or c = 2. However, 0 Ï (0,3). Hence the only such value is c = 2. Q7. Find the absolute maximum and minimum values of f ( x) = x - 2 sin x on [0,p ]. Solution f '( x) = 1 - 2 cos x p 1 Þ x = is a critical number in (0, p ). 4 2 We now compare the values of the function f at 0, p/4, and p: f '( x) = 0 Þ cos x = f (0) = 0 - 2 sin 0 = 0; f (p ) = p - 2 sin p = p ; and 1p æp ö p æp ö p f ç ÷ = - 2 sin ç ÷ = - 2 = - 1, which is a negative value. 24 è4ø 4 è4ø 4 \ Absolute maximum of the given function is f (p )=p p p and absolute minimum is f ( ) = - 1 4 4 3 Q8. Find the slope of the tangent line to the curve x3 + y 3 = 6 xy at the point (3,3). Solution Differentiating both sides implicitly with respect to x, we obtain: 3x...
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This document was uploaded on 02/19/2014 for the course MATH Math 113/1 at Grant MacEwan University.

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