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Unformatted text preview: Math 1a MVT and Two Derivative Tests Fall, 2009 1 Show that 1 2 x = 5 3 2 for some x between 3 and 5. 2 A differentiable function f ( x ) satisfies f ( x ) 2 for all x . If f (0) = 1, what can we say about f (3)? 3 Show that x + 1 1 + 1 2 x for x 0. 4 Suppose g ( x ) is differentiable for all x and g ( x ) is never 0. What is the maximum number of roots g ( x ) could have? 5 Explain why the cubic function x 3 5 x 2 + 3 x + 1 has exactly three roots. 6 Suppose f ( x ) is differentiable on [ a,b ] and f ( x ) > 0 for a x b . Explain why f ( a ) f ( b ). Some questions to answer about the following functions: (1) What is the domain of y = f ( x )? (2) Where is the tangent to y = f ( x ) vertical or horizontal? That is, what are the critical numbers of f ( x )? (3) Where is the graph of y = f ( x ) increasing or decreasing? (4) Where are the local maxima or local minima? Use the first derivative test to determine these points. (5) Where is the graph concave up? Where is it concave down? Where are the inflection points? Does the second derivative test confirm your values for the maima and minima? 7 f ( x ) = 9 x x 3 8 f ( x ) = x 3 + 6 x 2 + 12 x + 4 9 f ( x ) = x + 1 x = x 2 + 1 x 10 f ( x ) = ( x 2 9 ) 1 / 3 11 f ( x ) = 3 x 5 20 x 3 12 f ( x ) = x 2 + 1 x = x 3 + 1 x MVT and Two Derivative Tests Answers / Solutions 1 This is exactly the statement of the Mean Value Theorem (MVT) if we take f ( x ) = x , a = 3, b = 5, and c = x . Recall that the MVT says that there is a constant c in the interval ( a,b ) with f ( c ) = f ( b ) f ( a ) b a . In this case, f ( x ) = 1 2 x 1 / 2 = 1 2 x , so this says that there is an x in the interval (3 , 5) with f ( x ) = f (5) f (3) 5 3 or 1 2 x = 5 3 2 . Of course, we could solve this equation for x and find that x = 2 + 1 2 15 3 . 936492, but thats not really in the spirit of the problem. 2 The MVT in this case says that there is a c in the interval (0 , 3) with f (3) f (0) 3 = f ( c ) . Since f (0) = 1, this says that f (3) = 1 + 3 f ( c ). But we know that f ( c ) 2, so f (3) 1 + 3 2 = 7. 3 This is really just a statement about concavity. The function f ( x ) = x + 1 is concave down for x > 1 (essentially the entire domain). We can tell this because f 00 ( x ) = 1 4 ( x + 1) 3 / 2 is negative for all x > 1. This means that the tangent line to y = f ( x ) at ( x,y ) = (0 , 1) lies above the curve. It turns out that y = 1 + 1 2 x is this tangent line, so 1 + 1 2 x x + 1 for all x . Heres a quick sketch: ........................................................................................................................................................................................................................................................................................................ ....
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This note was uploaded on 03/27/2012 for the course MATH 1a taught by Professor Benedictgross during the Fall '09 term at Harvard.
 Fall '09
 BenedictGross
 Derivative

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