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Lesson 2: A Catalog of Essential Functions

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There are many classes of functions we use for modeling the world.

There are many classes of functions we use for modeling the world.

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  • 1. Section 1.2 A catalog of essential functions V63.0121.006/016, Calculus I January 21, 2010 Announcements Email Trang Nguyen (first name, @nyu.edu) with WebAssign enrollment issues Blackboard is acting up. See http://www.math.nyu.edu/ courses/Calculus/2010/Spring/121/syllabus.html . . . . . .
  • 2. Outline Modeling Classes of Functions Linear functions Other Polynomial functions Other power functions Rational functions Trigonometric Functions Exponential and Logarithmic functions Transformations of Functions Compositions of Functions . . . . . .
  • 3. The Modeling Process . . Real-world . . m . odel Mathematical . Problems Model s . olve .est t . i .nterpret . Real-world . Mathematical . Predictions Conclusions . . . . . .
  • 4. Plato’s Cave . . . . . .
  • 5. The Modeling Process . . Real-world . . m . odel Mathematical . Problems Model s . olve .est t . i .nterpret . Real-world . Mathematical . Predictions Conclusions S . hadows F . orms . . . . . .
  • 6. Outline Modeling Classes of Functions Linear functions Other Polynomial functions Other power functions Rational functions Trigonometric Functions Exponential and Logarithmic functions Transformations of Functions Compositions of Functions . . . . . .
  • 7. Classes of Functions linear functions, defined by slope an intercept, point and point, or point and slope. quadratic functions, cubic functions, power functions, polynomials rational functions trigonometric functions exponential/logarithmic functions . . . . . .
  • 8. Linear functions Linear functions have a constant rate of growth and are of the form f(x) = mx + b. . . . . . .
  • 9. Linear functions Linear functions have a constant rate of growth and are of the form f(x) = mx + b. Example In New York City taxis cost $2.50 to get in and $0.40 per 1/5 mile. Write the fare f(x) as a function of distance x traveled. . . . . . .
  • 10. Linear functions Linear functions have a constant rate of growth and are of the form f(x) = mx + b. Example In New York City taxis cost $2.50 to get in and $0.40 per 1/5 mile. Write the fare f(x) as a function of distance x traveled. Answer If x is in miles and f(x) in dollars, f(x) = 2.5 + 2x . . . . . .
  • 11. Example Biologists have noticed that the chirping rate of crickets of a certain species is related to temperature, and the relationship appears to be very nearly linear. A cricket produces 113 chirps per minute at 70 ◦ F and 173 chirps per minute at 80 ◦ F. (a) Write a linear equation that models the temperature T as a function of the number of chirps per minute N. (b) What is the slope of the graph? What does it represent? (c) If the crickets are chirping at 150 chirps per minute, estimate the temperature. . . . . . .
  • 12. Solution . . . . . .
  • 13. Solution The point-slope form of the equation for a line is appropriate here: If a line passes through (x0 , y0 ) with slope m, then the line has equation y − y0 = m(x − x0 ) . . . . . .
  • 14. Solution The point-slope form of the equation for a line is appropriate here: If a line passes through (x0 , y0 ) with slope m, then the line has equation y − y0 = m(x − x0 ) 80 − 70 10 1 The slope of our line is = = 173 − 113 60 6 . . . . . .
  • 15. Solution The point-slope form of the equation for a line is appropriate here: If a line passes through (x0 , y0 ) with slope m, then the line has equation y − y0 = m(x − x0 ) 80 − 70 10 1 The slope of our line is = = 173 − 113 60 6 So an equation for T and N is 1 1 113 T − 70 = (N − 113) =⇒ T = N − + 70 6 6 6 . . . . . .
  • 16. Solution The point-slope form of the equation for a line is appropriate here: If a line passes through (x0 , y0 ) with slope m, then the line has equation y − y0 = m(x − x0 ) 80 − 70 10 1 The slope of our line is = = 173 − 113 60 6 So an equation for T and N is 1 1 113 T − 70 = (N − 113) =⇒ T = N − + 70 6 6 6 37 If N = 150, then T = + 70 = 76 1 ◦ F 6 6 . . . . . .
  • 17. Other Polynomial functions Quadratic functions take the form f(x) = ax2 + bx + c The graph is a parabola which opens upward if a > 0, downward if a < 0. . . . . . .
  • 18. Other Polynomial functions Quadratic functions take the form f(x) = ax2 + bx + c The graph is a parabola which opens upward if a > 0, downward if a < 0. Cubic functions take the form f(x) = ax3 + bx2 + cx + d . . . . . .
  • 19. Example A parabola passes through (0, 3), (3, 0), and (2, −1). What is the equation of the parabola? . . . . . .
  • 20. Example A parabola passes through (0, 3), (3, 0), and (2, −1). What is the equation of the parabola? Solution The general equation is y = ax2 + bx + c. . . . . . .
  • 21. Example A parabola passes through (0, 3), (3, 0), and (2, −1). What is the equation of the parabola? Solution The general equation is y = ax2 + bx + c. Each point gives an equation relating a, b, and c: 3 = a · 02 + b · 0 + c −1 = a · 2 2 + b · 2 + c 0 = a · 32 + b · 3 + c . . . . . .
  • 22. Example A parabola passes through (0, 3), (3, 0), and (2, −1). What is the equation of the parabola? Solution The general equation is y = ax2 + bx + c. Each point gives an equation relating a, b, and c: 3 = a · 02 + b · 0 + c −1 = a · 2 2 + b · 2 + c 0 = a · 32 + b · 3 + c Right away we see c = 3. The other two equations become −4 = 4a + 2b −3 = 9a + 3b . . . . . .
  • 23. Solution (Continued) Multiplying the first equation by 3 and the second by 2 gives −12 = 12a + 6b −6 = 18a + 6b . . . . . .
  • 24. Solution (Continued) Multiplying the first equation by 3 and the second by 2 gives −12 = 12a + 6b −6 = 18a + 6b Subtract these two and we have −6 = −6a =⇒ a = 1. . . . . . .
  • 25. Solution (Continued) Multiplying the first equation by 3 and the second by 2 gives −12 = 12a + 6b −6 = 18a + 6b Subtract these two and we have −6 = −6a =⇒ a = 1. Substitute a = 1 into the first equation and we have −12 = 12 + 6b =⇒ b = −4 . . . . . .
  • 26. Solution (Continued) Multiplying the first equation by 3 and the second by 2 gives −12 = 12a + 6b −6 = 18a + 6b Subtract these two and we have −6 = −6a =⇒ a = 1. Substitute a = 1 into the first equation and we have −12 = 12 + 6b =⇒ b = −4 So our equation is y = x2 − 4x + 3 . . . . . .
  • 27. Other power functions Whole number powers: f(x) = xn . 1 negative powers are reciprocals: x−3 = 3 . x √ fractional powers are roots: x1/3 = 3 x. . . . . . .
  • 28. Rational functions Definition A rational function is a quotient of polynomials. Example x 3 (x + 3 ) The function f(x) = is rational. (x + 2)(x − 1) . . . . . .
  • 29. Trigonometric Functions Sine and cosine Tangent and cotangent Secant and cosecant . . . . . .
  • 30. Exponential and Logarithmic functions exponential functions (for example f(x) = 2x ) logarithmic functions are their inverses (for example f(x) = log2 (x)) . . . . . .
  • 31. Outline Modeling Classes of Functions Linear functions Other Polynomial functions Other power functions Rational functions Trigonometric Functions Exponential and Logarithmic functions Transformations of Functions Compositions of Functions . . . . . .
  • 32. Transformations of Functions Take the sine function and graph these transformations: ( π) sin x + ( 2 π) sin x − 2 π sin (x) + 2 π sin (x) − 2 . . . . . .
  • 33. Transformations of Functions Take the sine function and graph these transformations: ( π) sin x + ( 2 π) sin x − 2 π sin (x) + 2 π sin (x) − 2 Observe that if the fiddling occurs within the function, a transformation is applied on the x-axis. After the function, to the y-axis. . . . . . .
  • 34. Vertical and Horizontal Shifts Suppose c > 0. To obtain the graph of y = f(x) + c, shift the graph of y = f(x) a distance c units y = f(x) − c, shift the graph of y = f(x) a distance c units y = f(x − c), shift the graph of y = f(x) a distance c units y = f(x + c), shift the graph of y = f(x) a distance c units . . . . . .
  • 35. Vertical and Horizontal Shifts Suppose c > 0. To obtain the graph of y = f(x) + c, shift the graph of y = f(x) a distance c units upward y = f(x) − c, shift the graph of y = f(x) a distance c units y = f(x − c), shift the graph of y = f(x) a distance c units y = f(x + c), shift the graph of y = f(x) a distance c units . . . . . .
  • 36. Vertical and Horizontal Shifts Suppose c > 0. To obtain the graph of y = f(x) + c, shift the graph of y = f(x) a distance c units upward y = f(x) − c, shift the graph of y = f(x) a distance c units downward y = f(x − c), shift the graph of y = f(x) a distance c units y = f(x + c), shift the graph of y = f(x) a distance c units . . . . . .
  • 37. Vertical and Horizontal Shifts Suppose c > 0. To obtain the graph of y = f(x) + c, shift the graph of y = f(x) a distance c units upward y = f(x) − c, shift the graph of y = f(x) a distance c units downward y = f(x − c), shift the graph of y = f(x) a distance c units to the right y = f(x + c), shift the graph of y = f(x) a distance c units . . . . . .
  • 38. Vertical and Horizontal Shifts Suppose c > 0. To obtain the graph of y = f(x) + c, shift the graph of y = f(x) a distance c units upward y = f(x) − c, shift the graph of y = f(x) a distance c units downward y = f(x − c), shift the graph of y = f(x) a distance c units to the right y = f(x + c), shift the graph of y = f(x) a distance c units to the left . . . . . .
  • 39. Outline Modeling Classes of Functions Linear functions Other Polynomial functions Other power functions Rational functions Trigonometric Functions Exponential and Logarithmic functions Transformations of Functions Compositions of Functions . . . . . .
  • 40. Composition is a compounding of functions in succession g . ◦f . x . f . . g . . g ◦ f)(x) ( f .(x) . . . . . .
  • 41. Composing Example Let f(x) = x2 and g(x) = sin x. Compute f ◦ g and g ◦ f. . . . . . .
  • 42. Composing Example Let f(x) = x2 and g(x) = sin x. Compute f ◦ g and g ◦ f. Solution f ◦ g(x) = sin2 x while g ◦ f(x) = sin(x2 ). Note they are not the same. . . . . . .
  • 43. Decomposing Example √ Express x2 − 4 as a composition of two functions. What is its domain? Solution √ We can write the expression as f ◦ g, where f(u) = u and g(x) = x2 − 4. The range of g needs to be within the domain of f. To insure that x2 − 4 ≥ 0, we must have x ≤ −2 or x ≥ 2. . . . . . .