1 / 16

More Review for Test I

More Review for Test I. Calculation of Limits. Continuity. When you can calculate limits simply by ‘plugging in’. Geometric Definition : the function f is continuous at x = a if the graph of f is “not broken” at a. Analytic Definition: the function f is continuous at x = a if.

johnda
Download Presentation

More Review for Test I

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. More Review for Test I

  2. Calculation of Limits

  3. Continuity When you can calculate limits simply by ‘plugging in’ . Geometric Definition: the function f is continuous at x = a if the graph of f is “not broken” at a. Analytic Definition: the function f is continuous at x = a if = f(a)

  4. is continuous at x = 3 = f(3) is not continuous at x = 3 does not exist. Is continuous at 3 if W is chosen right f(3) = W + 3 = W + 3 = 9 They are equal exactly when W = 6

  5. Definition of the Derivative Geometric Definition: If f(x) is a function then f ‘ (a), the derivative of f at x= a is the slope of the tangent line to the graph of f(x) at (a, f(a)) Analytic or “limit” Definition

  6. Derivative Rules • For exam are responsible for: • (cf) ‘ = c f ‘ • (f + g) ‘ = f ‘ + g ‘ • (fg) ‘ = f ’g + f g ‘ • ( ) ‘ = • ( ) ‘ =

  7. Calculations with tangent lines 1. Tangent line to graph of f(x) at x = a General Eqn For Line m = f ‘(a) at x = 2 Eqn of tangent to graph of f ‘(2) = 7 a = 2, f(2) = -2, Eqn: y +2 = 7(x-2)

  8. Use of Rules: Suppose f(7) = 3 and f ‘(7) = 4. What is the equation of the tangent line to the graph of f(x) at x = 7? y – f(7) = f ‘(7) (x-7) y – 3 = 4 (x-7) What is the equation of the line through (7,3) which is perpendicular to the tangent line to the graph of f(x) at x = 7? y – 3 = -1/4 (x-7)

  9. Suppose f(9) =2, f ‘(9) = 4, g (9) = 8, g ‘(9) = 7. What is the equation of the tangent line to the graph ofh(x) = f(x) – g(x) at x = 9? y – h(9) = h ‘(9) (x-9) h(9) = f(9) – g(9) = 2-8 = -6 h ‘(9) = f ‘(9) – g ‘(9) = 4 – 7 = -3 y +6 = (-3)(x-9)

  10. Tangent Data From Graph What is the equation of the tangent line to the graph of h(x) = xf(x) at x = 3? y + 3 = h ‘(3)(x-3), h’(x) = f(x) + x f ‘(x). h ‘(3) = …

  11. Interpretation of the Tangent Line Philosophically, the tangent line to the graph of f(x) at (a, f(a)is the line that best approximates the graph of f “near(a,f(a))”

  12. Given f(x), and line y = mx + bhow can you tell if the lineis the tangent line at x = a? Answer:(1) y = f(a) when x = a ,(2) y ‘ (a) = f ‘(a) (since y ‘ (a) is the slope of the line and the slope of the line is f ‘(a)) Question: Is y = 9x -19 the tangent line to the graph of f(x) = at x = 4? Answer=NO When x = 4, y = 9*4-19 = 17 = f(4), but y ‘ (4) = 9 while f ‘(4) = 7

  13. Question: Suppose f ‘ (3) = 2 and f(3) = 5, g (3) =7 and g ‘(3) = 4. What is the equation of the tangent line to the graph of at x = 3? SLN: To know the tangent line at x =3 we need to know h(3) and h ‘(3) = y – h(3) = h ‘(3)*(x-3) is the equation

  14. The equation of the tangent line to the graph of f(x) at x = 5 isy = 14*x-32. What are f(5) and f ‘(5). Ans: The point (5, f(5)) is on the graph of f and on the tangent line.When x = 5, y = 14*5-32 = 38 so this has to be f(5).The slope of the line is y ‘= 14 so this has to be f ‘(5). The equation of the tangent line to the graph of h(x) = x*f(x)at x = 7 is y=5*x -3. What are f ‘(7) and f(7)?ANS: h ‘(x) = f(x) + x*f ‘(x) and y ‘(7) = 5 so 5 = h ‘(7) = f(7) + 7*f ‘(7).When x=7 y = 5*7-3=32 so 32 = h(7) = 7*f(7)Thus we have equations: 5 = f(7) + 7* f ‘(7) 32 = 7*f(7) which are easily solved for f(7) and f ‘(7).

  15. The derivative and Piecewise Functions If f is defined piecewise then f ‘(x) for x other than end points is calculated piecewise itself. Generallyif f is continuous at an end pointand the derivatives exist on the pieces and the limits of the derivatives exist (and are equal) from the left and the right then this is the derivative at the end points. If Except at x = 2 we calculate piecewise. f ‘(2) DNE since limit from the left is 4but limit from the right is -1

  16. Suppose Can C be chosen so that f ‘(x) has a derivative at x = 2?If f ‘(2) exists then f has to be continuous at x = 2. This meansthe limits from the right and left have to exist and be equal to f(2). = C8 = 3, so C must = 3/8 if f is to be continuous. From the left f ‘=3, from the right f ‘= 60/8 – f ‘DNE

More Related