Math1000 section2.4
- 1. Math 1000 Stuart Jones Section2.4 Average Rate of Change of Functions
- 2. Math 1000 Stuart Jones Theorem (Average Rate of Change of Functions) Let a function f (x) be given. The average rate of change from x = a to x = b is given by f (b) − f (a) b − a
- 3. Math 1000 Stuart Jones Theorem (Average Rate of Change of Functions) Let a function f (x) be given. The average rate of change from x = a to x = b is given by f (b) − f (a) b − a This is simply the slope of the secant line between those two points! (Secant line - a line that intersects a function at two points.)
- 4. Math 1000 Stuart Jones For each function, determine the average rate of change of the function between the given values of the variable. f (x)7x − 8 from x = 2 to x = 3
- 5. Math 1000 Stuart Jones For each function, determine the average rate of change of the function between the given values of the variable. f (x)7x − 8 from x = 2 to x = 3 h(t) = t2 + 3t from t = −1 to t = 5
- 6. Math 1000 Stuart Jones For each function, determine the average rate of change of the function between the given values of the variable. f (x)7x − 8 from x = 2 to x = 3 h(t) = t2 + 3t from t = −1 to t = 5 Let’s work out the second one.
- 7. Math 1000 Stuart Jones For each function, determine the average rate of change of the function between the given values of the variable. f (x)7x − 8 from x = 2 to x = 3 h(t) = t2 + 3t from t = −1 to t = 5 Let’s work out the second one. (5)2 + 3(5) − ((−1)2 + 3(−1)) 5 − (−1) 25 + 15 − (1 − 3) 6 40 + 2 6 42 6 = 7
- 8. Math 1000 Stuart Jones More Examples f (x) = x3 − 8x2 from x = 0 to x = 10
- 9. Math 1000 Stuart Jones More Examples f (x) = x3 − 8x2 from x = 0 to x = 10 Answer: 20 g(x) = 4 x from x = 5 to x = a Let’s work this one out. 4 a − 4 5 a − 5
- 11. Math 1000 Stuart Jones = 20−4a 5a a − 5 = 20 − 4a 5a · 1 a − 5
- 12. Math 1000 Stuart Jones = 20−4a 5a a − 5 = 20 − 4a 5a · 1 a − 5 = 20 − 4a 5a2 − 25a
- 13. Math 1000 Stuart Jones = 20−4a 5a a − 5 = 20 − 4a 5a · 1 a − 5 = 20 − 4a 5a2 − 25a = 4(5 − a) 5a(a − 5)
- 14. Math 1000 Stuart Jones = 20−4a 5a a − 5 = 20 − 4a 5a · 1 a − 5 = 20 − 4a 5a2 − 25a = 4(5 − a) 5a(a − 5) = −4(a − 5) 5a(a − 5)
- 15. Math 1000 Stuart Jones = 20−4a 5a a − 5 = 20 − 4a 5a · 1 a − 5 = 20 − 4a 5a2 − 25a = 4(5 − a) 5a(a − 5) = −4(a − 5) 5a(a − 5) = −4 5a
- 16. Math 1000 Stuart Jones This is related to the difference quotient also. A function is given. Determine the average rate of change of the function between the given values of the variable. f (x) = 3x2 from x = 4 to x = 4 + h
- 17. Math 1000 Stuart Jones This is related to the difference quotient also. A function is given. Determine the average rate of change of the function between the given values of the variable. f (x) = 3x2 from x = 4 to x = 4 + h Answer: 3h + 24
- 18. Math 1000 Stuart Jones The Bottom Line The Average Rate of Change formula is simply the slope of the secant line through two points. It is related to difference quotients and will be useful in calculus also. The most important takeaway for this section is to KNOW how to work with fractions (common denominators, etc.) and how to simplify.