L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 27. L43-Mon-12-Dec-2016-Rev-Cpt-4-HW44-and-Rev-Cpt-5-for-Final-HW45

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1 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 7 L43-Mon-1-Dec-016-Rev-Cpt-4-HW44-and-Rev-Cpt-5-for-Final-HW45

2 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 8 1. y-intercept: f f 0 -intercepts: 3 4 f so -intercepts at 0, 3, and 4. Multiplicities: 0, multiplicity, which is even so graph TOUCHES -ais here 3, multiplicity 1, which is odd so graph CROSSES -ais here 4, multiplicity 1, which is odd so graph CROSSES -ais here n 3. End behavior: Graph looks like f a. 4 As, f, which looks like y. 4. Ma number of turning points = n 1 = 4-1 = Behavior near each intercept: Near = -4, f 3 4 f j variable f 3 4 Near = 0, f g variable 3 4 f Near = 3, f h variable 4

3 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 9 Plot some specific points if necessary: f f

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6 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 3 1. Factor, reduce, find holes: Domain: 0, 5 Reduce: R Q Since the 's cancel, there will be a hole at = 0. To find the y-value of the hole, find Q So, there is a hole at (0, -.4).. Intercepts -intercepts: These are the values of in numerator that make function 0. Q or 3

7 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 33 = 4 has multiplicity 1, which is odd, so graph CROSSES here Calculate behavior near = 4: Q Variable Q 4 g = -3 has multiplicity 1, which is odd, so graph CROSSES here Calculate behavior near = -3: Q Variable Q 3 h y-intercept This is R(0) but 0 is not in domain so there is no y-intercept. There is a hole at (0, -.4) 3. Vertical asymptotes: These are values of in denominator that make it zero. The multiplicity of = -5 is odd, so near 5 the graph will go up on one side and down on other. 4. Horizontal or oblique asymptotes: Since the degree of the numerator is eactly 1 more than the degree of the denominator, there is an oblique asymptote. Use long division to find it OA is y =

8 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page Plot where graph crosses HA or OA by solving Function oblique asymptote Q y No solution so graph does not cross asymptote. 6. Plot a point or two to help with the sketch. Q 8 Q

9 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page Factor, reduce to find holes: There is a hole at = -. This implies a factor of + in both numerator and denominator.. Intercepts: -int at = -7 crosses suggests ( + 7) in numerator. -int at = 5 crosses suggests ( - 5) in numerator. 7 5 y-intercept is at about y = 1.5 We can check this later. 3. Vertical Asymptote: Caused by zeros in denominator. VA of = -4 suggests ( + 4) in denominator. VA of = 3 suggests ( - 3) in denominator Horizontal or Oblique Asymptote: Caused by relative degrees of numerator and denominator. HA of y = ½ suggests the degree of numerator and denominator are the same and the ratio of the coefficients of the dominant terms is ½ Does function cross HA or OA? We cannot tell from the graph. Check y-intercept: f Checks. Check degrees of numerator and denominator. They are the same (degree 3). Checks Check if function crosses HA.

10 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page f HA So, the graph crosses the HA when = 3. We cannot tell from the given graph if that is so but it certainly looks possible. Graphing our function and then zooming in around = 3 shows that the function does cross the HA there (hard to see!).

11 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 37 Put everything on one side and zero on the other and figure out the signs: The zeros of all the factors are: 1and 5 and 3 Interval: Try Signs of 5 P NN P NP P PP N PP Sign of f P N P N f 0? yes no yes no, 5 1,3 Note that = -5 and = -1 cause division by 0 so they are not included in solution. = 3 is OK since 0 in the numerator is allowed because this is not a strict inequality. So, the solution is

12 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page f g f g f 1 1 Domain f: Domain g: Domain So Domain of f g must respect domains of g and Note also that since 1 for f() we can say. So, we get 0, g so so Echange and y and solve for y. f : y 1 f : 3 4 3y 4 y y 3y 4 y 3y 4 y 3y 4 y 3 4 y 4 3 So, the inverse function is f 1 4 3

13 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 39 Domain of f is or,, 1 y y 3 or, 3 3, Range of f Domain of f, which is 3 so range is

14 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 40 We can write the eponential functions with the same base and then equate the eponents: or 4 These are both in the domain of the original equation. Or, if we could not write the epressions with the same bases, we would use logs:

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16 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 4 Put all logs together on one side of equation and then change to eponential: log5 3 1log5 1 log log5 3 log or -4 is not in the domain of the original equation so the only solution is =.

17 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 43 It is tempting to take the log of both sides in order to get the out of the eponent. ln 81 ln 3 9 We can break up the right side using the product property but we cannot break up the left side. ln 81 ln 3 ln 9 ln 81 ln 3 ln 9 Can we make the bases of the eponentials the same? Notice that 81 can be written as 4 (We could also use a base 3 since 9 3 and That would work but it would involve solving a degree 4 equation, which is more difficult, but doable in this case since it would be reducible to a quadratic equation.) Now, we have 9 and 9 so try a u substitution: u u u 9 39 Let u 9 u 3u 3u u or u 1 Now, back substitute: u 1 u ln 9 ln 0 ln 9 ln ln ln

18 L43-Mon-1-Dec-016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 44