Topics Section 2 1 Derivatives and Rate of Change Any errors you can nd in the solutions can be reported here and are greatly appreciated https forms gle rGXwBeet5a3c3kF6A 1 Let f be a function Find the equation of the slope of the secant line that passes through the points a f a and a h f a h Solution By secant line we re just emphasizing that the line is related to the function f x itself We carry out the normal procedure for nding the slope of the line that passes through two given points f a h f a a h a f a h f a h Read back to yourself how this explains why the derivative is the slope of a tangent line 2 Let f be a function Find the equation of the slope of the secant line that passes through the points a f a and b f b Solution 3 Let f x 2x2 f b f a b a a Find the slope of the line through the points 1 f 1 and 2 f 2 Solution We have 1 f 1 1 2 and 2 f 2 2 8 The slope of the line passing through these two points is 6 b Find the slope of the line through the points a f a and b f b Solution f b f a b a 2b2 2a2 b a 2 b a b a b a 2 b a c Compute lim b 1 f b f 1 b 1 Solution Substituting a 1 into the formula we just computed f b f 1 b 1 lim b 1 lim b 1 2 b 1 2 1 1 4 d Write the equation of the line tangent to f x 2x2 at x 1 Solution Part c above tells us the tangent slope is equal to 4 The tangent line has slope 4 and passes through the point 1 f 1 1 2 Therefore using the point slope equation of a line we nd the equation of the tangent line at x 1 is T x 4 x 1 2 1 e Sketch the function f x 2x2 and the tangent line to the function at x 1 Solution Notice the function just touches the graph of f x at x 1 4 Use the de nition of the derivative to nd the derivative of the function f x 3x2 4 at the point x 2 Solution We insert this speci c function f x into the de nition of the derivative carefully f cid 48 2 lim h 0 f 2 h f 2 cid 0 3 2 h 2 4 cid 1 cid 0 3 2 2 4 cid 1 h 3 4 4h h2 4 16 h lim h 0 lim h 0 lim h 0 lim h 0 12 h 12h 3h2 h 12 3h 2 5 Use the de nition of the derivative to nd the derivative function f cid 48 x corresponding to f x 1 x 2 Now what is f cid 48 1 f cid 48 10 Solution Notice the technique of nding a common denominator when two rational functions are added together cid 19 x h 2 x h 2 x 2 f cid 48 x lim h 0 1 1 h h cid 19 1 f x h f x x h 2 1 cid 18 x 2 cid 18 x h 2 x 2 x 2 x h 2 x h 2 x 2 x h 2 x h 2 x 2 1 h 1 h 1 h 1 h x 2 x 2 h 1 x h 2 x 2 1 lim h 0 lim h 0 lim h 0 lim h 0 lim h 0 lim h 0 x 2 x 2 1 x 2 2 Now we have a formula we can use f cid 48 x 1 x 2 2 We have f cid 48 1 1 6 Use the de nition of the derivative to nd the derivative f cid 48 6 where f x 1 2 2 1 x 4 9 And f cid 48 10 1 10 2 2 1 64 Solution 6 4 cid 33 2 2 2 h 2 h f cid 48 6 lim h 0 h f 6 h f 6 cid 112 6 h 4 2 h cid 32 2 h h 2 h 2 h 2 h 2 2 h 1 2 h 2 h 2 lim h 0 lim h 0 lim h 0 lim h 0 lim h 0 1 2 2 3 7 Using the de nition of the derivative compute g cid 48 1 where g x x2 3x Answer 1 Same method as the last few problems 8 Consider the function y 3 2 x point 1 3 a Using the de nition of the derivative compute the slope of the tangent line to the function y at the 3 2 x Solution This means computing y cid 48 1 The answer is y cid 48 1 3 b Find the equation of the line tangent to y x at x 1 Solution Using the tangent slope y cid 48 1 3 and the point slope formula the tangent line is T x 3 x 1 3 9 Let f x x3 3x 1 Using the de nition of the derivative compute the slope of the tangent line at the point a f a Where is the tangent line horizontal Use this to sketch a graph of y f x Partial solution When you re inputting into the de nition of the derivative f cid 48 a carefully you ll have a term a h 3 You will expand that out as a h 3 a h 2 a h a2 2ah h2 a h a3 2a2h ah2 a2h 2ah2 h3 a3 3a2h 3ah2 h3 Proceed carefully as before The answer is f cid 48 a 3a2 3 10 Suppose the position of a car is given by the function s t t t2 for t 0 a Find the average velocity of the car from t 0 to t 1 2 b Find the instantaneous velocity of the car at time t 1 c At what time is the car stopped Partial solution a The average velocity over this interval is v 1 2 v 0 1 2 0 1 4 0 1 2 1 2 b The instantaneous velocity at t 1 is s cid 48 1 which if you compute carefully you will nd is equal to 1 c Being stopped means instantaneous velocity is equal to 0 The car is stopped at t 1 and computing the derivative yields s cid 48 t 1 2t at any time t Therefore solving 1 2t 0 we see that at t 1 2 we have s cid 48 t 0 The car is stopped at t 1 2 4 11 Show that r t is not di erentiable at t 0 What does it mean for r t to not be di erentiable cid 40 1 t 1 t …
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