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Let’s go back in time …. Unit 3: Derivative Applications

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1 Let’s go back in time …. Unit 3: Derivative Applications
Unit 2: The Derivative lim π‘₯β†’π‘Ž 𝑓(π‘₯) Unit 1: The LIMIT of a Function This is the process of looking at what is happening to a function at certain values of π‘₯, in particular what value, if any, is 𝑓(π‘₯) approaching as π‘₯ approaches π‘Ž.

2 4.3 Vertical and Horizontal Asymptotes
Vertical Asymptotes of Rational Functions A rational function of the form 𝑓 π‘₯ = 𝑝(π‘₯) π‘ž(π‘₯) has a vertical asymptote π‘₯=𝑐 if π‘ž 𝑐 =0 AND 𝑝(𝑐)β‰ 0. OK … but can we make a connection to limits and calculus?

3 4.3 Vertical and Horizontal Asymptotes
Vertical Asymptotes and Infinite Limits The graph of 𝑓 π‘₯ has a vertical asymptote, π‘₯=𝑐, if one (or more) of the following is true: lim π‘₯β†’ 𝑐 βˆ’ 𝑓 π‘₯ =±∞ lim π‘₯β†’ 𝑐 + 𝑓 π‘₯ =±∞

4 Determine any vertical asymptotes of the function 𝑓 π‘₯ = π‘₯ π‘₯ 2 +π‘₯βˆ’2 .
Example #1: 𝒙→ βˆ’πŸ βˆ’ 𝒙→ βˆ’πŸ + 𝒙→ 𝟏 βˆ’ 𝒙→ 𝟏 + Sign of 𝑓(π‘₯) 𝑓 π‘₯ β†’? 0= π‘₯ 2 +π‘₯βˆ’2 0=(π‘₯+2)(π‘₯βˆ’1) π‘₯=βˆ’2, 1 Note: neither of these values make the numerator zero β†’ vertical asymptotes occur there. BUT … how do you know if the graph is heading towards positive or negative infinity? We need this information to sketch the curve.

5 4.3 Vertical and Horizontal Asymptotes
lim π‘₯β†’βˆ’βˆž 𝑓 π‘₯ lim π‘₯β†’+∞ 𝑓 π‘₯ Limits and End Behaviour Notation: lim π‘₯β†’Β±βˆž 1 π‘₯ =0

6 Let’s go back in time …. AGAIN
What algebraic techniques did we use to find the limit of a function? Direct Substitution Factoring Rationalizing

7 Algebraic technique Warmup: Write each function so the term of highest degree is a factor. (a) 𝑝 π‘₯ = π‘₯ 2 +4π‘₯+1 (b) π‘ž π‘₯ = 3π‘₯ 3 βˆ’4π‘₯+5 = π‘₯ π‘₯ + 1 π‘₯ 2 = 3π‘₯ π‘₯ π‘₯ 3 Why would this technique be useful for determining the end behavior of a function?

8 Algebraic technique Determine the value of each of the following:
(a) lim π‘₯β†’+∞ 2π‘₯βˆ’3 π‘₯+3 = 2 1βˆ’ = lim π‘₯β†’+∞ 2π‘₯ 1βˆ’ 3 2π‘₯ π‘₯ 1+ 1 π‘₯ =2 = lim π‘₯β†’+∞ 2 1βˆ’ 3 2π‘₯ π‘₯

9 Determine the value of each of the following:
(b) lim π‘₯β†’βˆ’βˆž π‘₯ π‘₯ 2 +1 = 1 lim π‘₯β†’βˆ’βˆž π‘₯ 1+0 = lim π‘₯β†’βˆ’βˆž π‘₯ π‘₯ π‘₯ 2 = lim π‘₯β†’βˆ’βˆž 1 π‘₯ =0 = lim π‘₯β†’βˆ’βˆž 1 π‘₯ π‘₯ 2 = 1 lim π‘₯β†’βˆ’βˆž π‘₯ lim π‘₯β†’βˆ’βˆž π‘₯ 2

10 Determine the value of each of the following:
(c) lim π‘₯β†’+∞ 2π‘₯ π‘₯ 2 βˆ’π‘₯+4 = 2+0 3βˆ’0+0 = 2 3 = lim π‘₯β†’+∞ π‘₯ π‘₯ π‘₯ 2 3βˆ’ 1 π‘₯ + 4 π‘₯ 2 = lim π‘₯β†’+∞ π‘₯ 2 3βˆ’ 1 π‘₯ + 4 π‘₯ 2

11 4.3 Vertical and Horizontal Asymptotes
Horizontal Asymptotes and Limits at Infinity If lim π‘₯β†’+∞ 𝑓 π‘₯ =𝐿 and/or lim π‘₯β†’βˆ’βˆž 𝑓 π‘₯ =𝐿 , we say that the line 𝑦=𝐿 is a horizontal asymptote of the graph 𝑓(π‘₯).

12 Example: Determine the equation of any horizontal asymptotes of the function 𝑓(π‘₯)= 3π‘₯+5 2π‘₯βˆ’1 . State whether the graph approaches the asymptote from above or below. lim π‘₯β†’+∞ 𝑓(π‘₯) = lim π‘₯β†’+∞ (3π‘₯+5) lim π‘₯β†’+∞ (2π‘₯βˆ’1) Do we have enough information to sketch the end behaviour of 𝑓 π‘₯ ? Solution: NO! We need to know if the graph is approaching this asymptote from above or below. = lim π‘₯β†’+∞ 3+ 5 π‘₯ lim π‘₯β†’+∞ 2βˆ’ 1 π‘₯ = 3 2 Similarly, we can show that lim π‘₯β†’βˆ’βˆž 𝑓(π‘₯) = 3 2 →𝑦= 3 2 is a horizontal asymptote.

13 Example: Determine the equation of any horizontal asymptotes of the function 𝑓(π‘₯)= 3π‘₯+5 2π‘₯βˆ’1 . State whether the graph approaches the asymptote from above or below. Solution: How can we do this? Pick a very large positive and negative value for π‘₯ and evaluate 𝑓(π‘₯) to see whether it is larger or smaller (above or below) 𝑦= 3 2 . 𝑓 1000 = βˆ’1 = > 3 2 𝑓 βˆ’1000 = 3 βˆ’ βˆ’1000 βˆ’1 = βˆ’2995 βˆ’2001 < 3 2 Therefore, as π‘₯β†’+∞, 𝑓 π‘₯ β†’y= 3 2 from above and as π‘₯β†’βˆ’βˆž, 𝑓 π‘₯ β†’y= 3 2 from below.

14 Lastly … Oblique Asymptotes
In a rational function, and oblique asymptote occurs when the degree of the numerator is exactly one greater than the degree of the denominator. You can find the oblique asymptote by dividing the numerator by the denominator. The first two terms of the quotient are the oblique asymptote.

15 Example: Determine the equation of the oblique asymptote of the function 𝑓(π‘₯)= 2 π‘₯ 2 +3π‘₯βˆ’1 π‘₯+1 . State whether the graph approaches the asymptote from above or below. Solution: βˆ’1 βˆ’1 ↓ βˆ’2 βˆ’1 2 1 βˆ’2 Therefore , the line π’š=πŸπ’™+𝟏 is an oblique asymptote of the curve.

16 In summary … to sketch a curve, apply these steps …
Check for any discontinuities in the domain (values of π‘₯ that make the denominator zero). Determine if there are vertical asymptotes at these discontinuities (any of these π‘₯’s that don’t make the numerator zero as well), and determine the direction from which the curve approaches these asymptotes (i.e. does 𝑓 π‘₯ β†’Β±βˆž?). Find any 𝒙 𝐨𝐫 π’š intercept(s). Find any critical points (values of π‘₯ for which 𝑓 π‘₯ =0 or 𝑓 π‘₯ is undefined). Create an interval of increase/decrease and test values in the derivative function in each interval to determine if the function is increasing or decreasing. Use this knowledge to determine if there is a local or absolute extrema (max/min) at each critical point. Test the end behaviour to find any horizontal asymptotes by determining lim π‘₯β†’Β±βˆž 𝑓(π‘₯) . Evaluate the function at a positive and negative value of π‘₯ to determine whether the function is approaching the asymptote from above or below. If the degree of the numerator is exactly one degree more than the denominator, then divide the numerator by the denominator. The first two terms of the quotient is the equation of the oblique asymptote.

17 QUESTIONS: p.193-195 #1, 2, 3ac, 4bdf, 5cd, 7bd, 10


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