chapter4ProblemsAndSolutions

# chapter4ProblemsAndSolutions - Chapter 4 The Derivative 4.1...

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Unformatted text preview: Chapter 4 The Derivative 4.1 You are given the following information about the signs of the derivative of a function, f ( x ). Use this information to sketch a (very rough) graph of the function for- 3 < x < 3. x-3-2-1 1 2 3 f prime ( x ) +- + + Detailed Solution: See Figure 4.1.-2 -1 0 1 2 3 4 x f(x)-2 -1 0 1 2 3 4 Figure 4.1: Figure for Problem 4.1 4.2 You are given the following information about the the values of the derivative of a function, g ( x ). Use this information to sketch (very rough) graph the function for- 3 < x < 3. x-3-2-1 1 2 3 g prime ( x )-1 2 1-1-2 v.2005.1 - September 4, 2009 1 Math 102 Problems Chapter 4 Detailed Solution: A rough sketch of the desired function is shown in Figure 4.2. x-3 -2 -1 0 1 2 3 4-3 -2 -1 0 1 2 3 4 g(x) Figure 4.2: Figure for Problem 4.2 4.3 What is the slope of the tangent line to the function y = f ( x ) = 5 x + 2 when x = 2? when x = 4 ? How would this slope change if a negative value of x was used? Why? Detailed Solution: The equation y = 5 x +2 describes a straight line with slope 5. The slope is the same at every point, and the tangent line at each point is the same as the original line. Thus, the slope of the tangent line for any value of x is 5 . 4.4 Find the equation of the tangent line to the function y = f ( x ) = | x + 1 | at: (a) x =- 1, (b) x =- 2, (c) x = 0. If there is a problem finding a tangent line at one of these points, indicate what the problem is. Detailed Solution: The graph of the function is shown in Figure 4.3. It consists of two straight line with slopes 1 and-1, joined at a cust (at x =- 1). Those lines are y = ( x + 1) and y =- ( x + 1). (a) This is the cusp, and hence the derivative does not exist here and there is no tangent line. v.2005.1 - September 4, 2009 2 Math 102 Problems Chapter 4 x y-1 y=| x+1 | Figure 4.3: Figure for Problem 4.4 (b) The slope at x =- 2 is -1, and so is the derivative. The tangent line is the same as the straight line that forms the branch of | x + 1 | at this location, i.e. y =- ( x + 1) (b) The slope and the derivative at x = 0 is 1. The tangent line is y = ( x + 1) here, by similar reasoning. 4.5 A function f ( x ) has as its derivative f prime ( x ) = 2 x 2- 3 x (a) In what regions is f increasing or decreasing? (b) Find any local maxima or minima. (c) Is there an absolute maximum or minimum value for this function? Detailed Solution: See Figure 4.4 for a sketch of the derivative, f prime ( x ), and for an indication of where it is positive and negative (i.e. below or above the x axis).-1 1 2 3 4 5-1 1 2 x fâ(x)=2x^2-3x Figure 4.4: Figure for solution to Problem 4.5 We see from the sketch that v.2005.1 - September 4, 2009 3 Math 102 Problems Chapter 4 (a) f prime ( x ) = 2 x 2- 3 x < 0, which means that f is decreasing for 0 < x < 1 . 5. Also, f prime ( x ) = 2 x 2- 3 x > 0 which means that f is increasing for-â < x < , 1 . 5 < x < â ....
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## This note was uploaded on 05/19/2011 for the course MATH 102 Math 102 taught by Professor Allard during the Winter '09 term at UBC.

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chapter4ProblemsAndSolutions - Chapter 4 The Derivative 4.1...

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