Math 221 Section 2 2 Derivative as a Function Solutions Instructions Listen to your TA s instructions There are substantially more problems on this worksheet than we expect to be done in discussion and your TA might not have you do problems in order The worksheets are intentionally longer than will be covered in discussion in order to give students additional practice problems they may use to study Do not worry if you do not finish the worksheet 1 Let f x 2x2 x 1 a Compute the derivative f x using the definition of the derivative f x h f x 2 x h 2 x h 1 2x2 x 1 lim h 0 2 x2 2xh h2 x h 1 2x2 x 1 lim h 0 h lim h 0 h h lim h 0 2x2 4xh 2h2 h 2x2 h 4xh 2h2 h lim h 0 h lim h 0 4x 1 2h 4x 1 b Your friend says that the equation for the tangent line to f x at the point 1 4 is y 4 4x 1 x 1 What did they do wrong At the point 1 4 the value of f x is f 1 5 not 4x 1 The slope of the tangent line is equal to the value of the derivative at that point which is 5 The correct answer is y 4 5 x 1 2 Use the definition of the derivative to find the derivative f x where f x x 1 x h 1 x 1 lim h 0 h x h 1 x 1 x 1 h x h 1 lim h 0 h h x h 1 x 1 x h 1 x 1 h f x h f x h x 1 x 1 lim h 0 x h 1 x h 1 lim h 0 lim h 0 1 x h 1 x 1 2 1 x 1 3 Let f x x x What is f c for c 0 What is f c for c 0 What about f 0 For x 0 x x so f x x x x x 2x Then for x 0 2 x h 2x f x h f x lim h 0 h lim h 0 h For x 0 x x for f x x x x x 0 Then for x 0 lim h 0 2h h 2 At x 0 we have since if h 0 then 0 h 0 and f x h f x lim h 0 h lim h 0 0 h 0 f 0 h f 0 lim h 0 h lim h 0 2 0 h 2 h 0 h f 0 h f 0 lim h 0 lim h 0 0 h f x h f x h 1 since if h 0 then 0 h 0 Then limh 0 agree so f 0 does not exist does not exist because the sided limits do not 4 Is the function continuous at x 0 Is it differentiable at x 0 Then f x is continuous at x 0 because if h 0 then 0 h 0 cid 26 0 x2 x 0 x 0 f x lim x 0 f x lim f x lim lim x 0 x 0 x 0 0 0 x2 0 f 0 02 0 f 0 h f 0 0 0 h 0 lim h 0 lim h 0 f 0 h f 0 0 h 2 0 lim h 0 h lim h 0 lim h 0 lim h 0 h 0 h2 h h h cid 26 ax2 b x x2 f x x 1 x 1 Then limh 0 f 0 h f 0 h exists so f x is differentiable at x 0 5 For which values of a and b is the following function differentiable at x 1 Sketch a graph for those values of a and b For f x to be differentiable at x 1 it must be continuous at x 1 and so limx f x limx 1 ax2 b a b limx 1 f x 1 12 0 Then a b 0 so b a Now f 1 h f 1 lim h 0 h a 1 h 2 b 0 lim h 0 a a 2ah ah2 h a 2ah ah2 b lim h 0 h 2ah ah2 lim h 0 lim h 0 a b 2ah ah2 h f 1 h f 1 h lim h 0 lim h 0 1 h 1 h 2 0 h lim h 0 1 h 1 2h h2 h 0 2a ah 2a h h2 lim h h lim h 0 h h lim h 0 1 h 1 Then 2a 1 so a 1 lim h 0 2 and since b a b 1 2 2 6 Suppose f x is a function which passes through the point 4 3 and that the line tangent to y f x at 4 3 also passes through the point 0 2 a Sketch the tangent line along with a possible graph of f x make sure to label the two given points b Find an equation of the tangent line you drew The tangent line passes through the points 4 3 and 0 2 so its slope is m 3 2 its equation is y 1 Then the equation is y 1 4 x b for some b Since it passes through 0 2 we have 2 1 4 0 1 4 Then 4 0 b so b 2 4 x 2 c What is f 4 What is f 4 The graph of f x passes through 4 3 so f 4 3 The slope of the tangent line is 1 4 so f 4 1 4 3
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