Chapter 2. Exercise 5. Evaluate the limit

Transcription

1 Chapter Exercise 5 Evaluate the limit f x_ : x 9 x x As x - this produces a 0 in both the numerator and denominator, which means we should factor and simplify before we evaluate. num Factor Numerator f x x x denom Factor Denominator f x x x ratio Simplify num denom x x ratio. x Exercise 7 Evaluate the limit. This also goes to 0 in the numerator and denominator, but here we merely need to expand algebraically and simplify by cancellation ratio Expand h h h h At this point we can evaluate with h = 0 ratio. h 0 Exercise 9 Evaluate the limit. Only the denominator goes to zero as r 9, so basically the limit does not exist (or more technically, approaches, an example of a vertical asymptote).

2 review.nb r ratio r 9 4 r 9 r 4 ratio. r 9 ComplexInfinity Exercise 4 Find the limit as x -. Note that numerator and denominator will both be very negative, to produce a positive result. f x_ : x x 4 5 x x 4 num x ^ 4 Expand Numerator f x x ^ 4 x 4 x x 4 denom x ^ 4 Expand Denominator f x x ^ 4 5 x 4 x x 4 num denom x 4 x 5 x 4 x The reciprocal powers of x drop out Limit num denom, x Exercise 5 Find the limit as x Note that numerator and denominator will both be very large positive, to produce a positive result. f x_ : x 9 x 6

3 review.nb num x Map x &, Numerator f x, 9 x x denom x Expand Denominator f x x 6 x x num denom 9 x 6 x The reciprocal powers of x drop out Limit num denom, x Exercise f x_ : x x 0 x 0 x x x Pick the appropriate portion of the function depending upon where you are approaching the point from Limit f x, x 0, Direction From the right Limit f x, x 0, Direction From the right 0 Since they don't agree, the overall limit does not exist Limit f x, x, Direction 0 Limit f x, x, Direction 0 Since they agree the overall limit is 0 as well

4 4 review.nb Exercise 7 Clear s s t_ : t t 4 Average velocity is change in s divided by change in t s s s s 4 s.5 s.5.65 s. s..55

5 review.nb 5 Exercise 9 4 In increasing order f '' 5., 0, f ' 5., f '.,, f ' , 0, 0.649, ,,.08 Exercise 0 f x_ : x x Find derivative at x = ratio 4 x x x f x f x num Factor Numerator ratio x x x ratio Simplify num Denominator ratio x x Derivative is the slope of the tangent line m ratio. x 0 Using point-slope formula through (, 4) tangent y 4 m x 4 y 0 x

6 6 review.nb Reduce tangent, y y 6 0 x Plot f x, 6 0 x, x, 0, Exercise f x_ : x rightratio f.0 f.0 leftratio f.0 f.0 m Mean leftratio, rightratio tangent y f m x y x Reduce tangent, y y x

7 review.nb 7 Plot f x, x, x, 0, Exercise 5 Given f Sketch f' Exercise 7 f x_ : 5 x

8 8 review.nb Find the derivative f'[x] ratio f x h f x h 5 x 5 h x h num Expand Numerator ratio 5 x 5 h x 5 h denom Denominator ratio 5 x 5 h x h 5 x 5 h x ratio Simplify num denom 5 5 x 5 h x deriv ratio. h x Find the domain of f[x] Reduce 5 x 0, x x 5 Find the domain of f'[x] (recall the denominator cannot be equal zero) Reduce 5 x 0, x x 5 Plot f x, deriv, x, 0,

9 review.nb 9 Exercise 9 Not differentiable at x = -4 (not continuous, break), x = - (cusp), x = (not continuous, infinite), x = 5 (vertical tangent) Exercise 40 Can see that c is positive where a is increasing and negative when a is decreasing. Also can see that b is positive when c is increasing and negative when c is decreasing. So f = a, f' = c, and f'' = b. Exercise 4 data 980, 9.9, 985, 87., 990, 7.9, 995, 409., 000, ; Estimate the derivative at 990 (rate of change in the amount of currency in circulation in billions of dollars per year) by averaging the secant slope before with the secant slope after m 7.48 m 6.9 data 4, data, data 4, data, data, data, data, data, estimate. m m Exercise 4 Given the following f' f is increasing when f' is positive (- < x < 0 or x > ). f is decreasing everywhere else. There will be a local maximum when f' goes from positive to negative at x = 0. There will be a local minimum when f' goes from negative to positive at x = - and x =. f is concave up when f' is increasing (x < - or x > ) and concave down when f' is decreasing (- < x < ). A sketch of f (the antiderivative of f') with f(0) = 0 is as follows.

10 0 review.nb Exercise 44 Here is an approximate graph of f'. Note that in the book the mins and maxs all fall at nice integer values, which I couldn't quite arrange for. 4 6 So for f'' the zeros should be at x = -, x =, x=, x =, and x = For f, the peaks should all occur at x = -, x = 0, x = (actually a horizontal point of inflection since f' doesn't change sign there), and x = 4, since that is where f' = 0.

11 review.nb 4 6 Exercise 45 It is probably best to start off with the derivatives to get direction, then introduce the second derivatives to get curvature, then identify the specific function values and the limits of the function. One possibility looks sort of like the following